SEVENTH EDITION
ELECTRONIC DEVICES
AND CIRCUIT THEORY
ROBERT BOYLESTAD
LOUIS NASHELSKY
PRENTICE HALL
Upper Saddle River, New Jersey
Columbus, Ohio
Contents
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
1.14
1.15
1.16
1.17
2
2.1
2.2
2.3
PREFACE
xiii
ACKNOWLEDGMENTS
xvii
SEMICONDUCTOR DIODES
1
Introduction 1
Ideal Diode 1
Semiconductor Materials 3
Energy Levels 6
Extrinsic Materials—n- and p-Type 7
Semiconductor Diode 10
Resistance Levels 17
Diode Equivalent Circuits 24
Diode Specification Sheets 27
Transition and Diffusion Capacitance 31
Reverse Recovery Time 32
Semiconductor Diode Notation 32
Diode Testing 33
Zener Diodes 35
Light-Emitting Diodes (LEDs) 38
Diode Arrays—Integrated Circuits 42
PSpice Windows 43
DIODE APPLICATIONS
51
Introduction 51
Load-Line Analysis 52
Diode Approximations 57
v
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
3.11
3.12
4
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
4.13
5
5.1
5.2
5.3
vi
Contents
Series Diode Configurations with DC Inputs 59
Parallel and Series-Parallel Configurations 64
AND/OR Gates 67
Sinusoidal Inputs; Half-Wave Rectification 69
Full-Wave Rectification 72
Clippers 76
Clampers 83
Zener Diodes 87
Voltage-Multiplier Circuits 94
PSpice Windows 97
BIPOLAR JUNCTION TRANSISTORS
112
Introduction 112
Transistor Construction 113
Transistor Operation 113
Common-Base Configuration 115
Transistor Amplifying Action 119
Common-Emitter Configuration 120
Common-Collector Configuration 127
Limits of Operation 128
Transistor Specification Sheet 130
Transistor Testing 134
Transistor Casing and Terminal Identification 136
PSpice Windows 138
DC BIASING—BJTS
143
Introduction 143
Operating Point 144
Fixed-Bias Circuit 146
Emitter-Stabilized Bias Circuit 153
Voltage-Divider Bias 157
DC Bias with Voltage Feedback 165
Miscellaneous Bias Configurations 168
Design Operations 174
Transistor Switching Networks 180
Troubleshooting Techniques 185
PNP Transistors 188
Bias Stabilization 190
PSpice Windows 199
FIELD-EFFECT TRANSISTORS
Introduction 211
Construction and Characteristics of JFETs 212
Transfer Characteristics 219
211
5.4
5.5
5.6
5.7
5.8
5.9
5.10
5.11
5.12
5.13
6
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
6.12
6.13
7
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
8
8.1
8.3
8.3
8.4
8.3
8.6
Specification Sheets (JFETs) 223
Instrumentation 226
Important Relationships 227
Depletion-Type MOSFET 228
Enhancement-Type MOSFET 234
MOSFET Handling 242
VMOS 243
CMOS 244
Summary Table 246
PSpice Windows 247
FET BIASING
253
Introduction 253
Fixed-Bias Configuration 254
Self-Bias Configuration 258
Voltage-Divider Biasing 264
Depletion-Type MOSFETs 270
Enhancement-Type MOSFETs 274
Summary Table 280
Combination Networks 282
Design 285
Troubleshooting 287
P-Channel FETs 288
Universal JFET Bias Curve 291
PSpice Windows 294
BJT TRANSISTOR MODELING
305
Introduction 305
Amplification in the AC Domain 305
BJT Transistor Modeling 306
The Important Parameters: Zi, Zo, Av, Ai 308
The re Transistor Model 314
The Hybrid Equivalent Model 321
Graphical Determination of the h-parameters 327
Variations of Transistor Parameters 331
BJT SMALL-SIGNAL ANALYSIS
338
Introduction 338
Common-Emitter Fixed-Bias Configuration 338
Voltage-Divider Bias 342
CE Emitter-Bias Configuration 345
Emitter-Follower Configuration 352
Common-Base Configuration 358
Contents
vii
8.7
8.8
8.9
8.10
8.11
8.12
8.13
9
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.10
9.11
9.12
9.13
9.14
9.15
10
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8
10.9
10.10
10.11
10.12
11
11.1
11.2
11.3
viii
Contents
Collector Feedback Configuration 360
Collector DC Feedback Configuration 366
Approximate Hybrid Equivalent Circuit 369
Complete Hybrid Equivalent Model 375
Summary Table 382
Troubleshooting 382
PSpice Windows 385
FET SMALL-SIGNAL ANALYSIS
401
Introduction 401
FET Small-Signal Model 402
JFET Fixed-Bias Configuration 410
JFET Self-Bias Configuration 412
JFET Voltage-Divider Configuration 418
JFET Source-Follower (Common-Drain) Configuration 419
JFET Common-Gate Configuration 422
Depletion-Type MOSFETs 426
Enhancement-Type MOSFETs 428
E-MOSFET Drain-Feedback Configuration 429
E-MOSFET Voltage-Divider Configuration 432
Designing FET Amplifier Networks 433
Summary Table 436
Troubleshooting 439
PSpice Windows 439
SYSTEMS APPROACH—
EFFECTS OF Rs AND RL
452
Introduction 452
Two-Port Systems 452
Effect of a Load Impedance (RL) 454
Effect of a Source Impedance (Rs) 459
Combined Effect of Rs and RL 461
BJT CE Networks 463
BJT Emitter-Follower Networks 468
BJT CB Networks 471
FET Networks 473
Summary Table 476
Cascaded Systems 480
PSpice Windows 481
BJT AND JFET FREQUENCY RESPONSE
Introduction 493
Logarithms 493
Decibels 497
493
11.4
11.5
11.6
11.7
11.8
11.9
11.10
11.11
11.12
11.13
12
12.1
12.2
12.3
12.4
12.5
12.6
12.7
12.8
12.9
12.10
12.11
General Frequency Considerations 500
Low-Frequency Analysis—Bode Plot 503
Low-Frequency Response—BJT Amplifier 508
Low-Frequency Response—FET Amplifier 516
Miller Effect Capacitance 520
High-Frequency Response—BJT Amplifier 523
High-Frequency Response—FET Amplifier 530
Multistage Frequency Effects 534
Square-Wave Testing 536
PSpice Windows 538
COMPOUND CONFIGURATIONS
Introduction 544
Cascade Connection 544
Cascode Connection 549
Darlington Connection 550
Feedback Pair 555
CMOS Circuit 559
Current Source Circuits 561
Current Mirror Circuits 563
Differential Amplifier Circuit 566
BIFET, BIMOS, and CMOS Differential Amplifier Circuits 574
PSpice Windows 575
13
DISCRETE AND IC
MANUFACTURING TECHNIQUES
13.1
13.2
13.3
13.4
13.5
13.6
13.7
13.8
13.9
Introduction 588
Semiconductor Materials, Si, Ge, and GaAs 588
Discrete Diodes 590
Transistor Fabrication 592
Integrated Circuits 593
Monolithic Integrated Circuit 595
The Production Cycle 597
Thin-Film and Thick-Film Integrated Circuits 607
Hybrid Integrated Circuits 608
14
14.1
14.2
14.3
14.4
14.5
14.6
14.7
14.8
544
OPERATIONAL AMPLIFIERS
588
609
Introduction 609
Differential and Common-Mode Operation 611
Op-Amp Basics 615
Practical Op-Amp Circuits 619
Op-Amp Specifications—DC Offset Parameters 625
Op-Amp Specifications—Frequency Parameters 628
Op-Amp Unit Specifications 632
PSpice Windows 638
Contents
ix
15
15.1
15.2
15.3
15.4
15.5
15.6
15.7
16
16.1
16.2
16.3
16.4
16.5
16.6
16.7
16.8
16.9
17
17.1
17.2
17.3
17.4
17.5
17.6
17.7
17.8
18
18.1
18.2
18.3
18.4
18.5
18.6
18.7
18.8
18.9
18.10
x
Contents
OP-AMP APPLICATIONS
648
Constant-Gain Multiplier 648
Voltage Summing 652
Voltage Buffer 655
Controller Sources 656
Instrumentation Circuits 658
Active Filters 662
PSpice Windows 666
POWER AMPLIFIERS
679
Introduction—Definitions and Amplifier Types 679
Series-Fed Class A Amplifier 681
Transformer-Coupled Class A Amplifier 686
Class B Amplifier Operation 693
Class B Amplifier Circuits 697
Amplifier Distortion 704
Power Transistor Heat Sinking 708
Class C and Class D Amplifiers 712
PSpice Windows 714
LINEAR-DIGITAL ICs
721
Introduction 721
Comparator Unit Operation 721
Digital-Analog Converters 728
Timer IC Unit Operation 732
Voltage-Controlled Oscillator 735
Phase-Locked Loop 738
Interfacing Circuitry 742
PSpice Windows 745
FEEDBACK AND OSCILLATOR CIRCUITS
Feedback Concepts 751
Feedback Connection Types 752
Practical Feedback Circuits 758
Feedback Amplifier—Phase and Frequency Considerations 765
Oscillator Operation 767
Phase-Shift Oscillator 769
Wien Bridge Oscillator 772
Tuned Oscillator Circuit 773
Crystal Oscillator 776
Unijunction Oscillator 780
751
19
19.1
19.2
19.3
19.4
19.5
19.6
19.7
20
20.1
20.2
20.3
20.4
20.5
20.6
20.7
20.8
20.9
20.10
20.11
21
21.1
21.2
21.3
21.4
21.5
21.6
21.7
21.8
21.9
21.10
21.11
21.12
21.13
21.14
21.15
21.16
POWER SUPPLIES
(VOLTAGE REGULATORS)
783
Introduction 783
General Filter Considerations 783
Capacitor Filter 786
RC Filter 789
Discrete Transistor Voltage Regulation 792
IC Voltage Regulators 799
PSpice Windows 804
OTHER TWO-TERMINAL DEVICES
810
Introduction 810
Schottky Barrier (Hot-Carrier) Diodes 810
Varactor (Varicap) Diodes 814
Power Diodes 818
Tunnel Diodes 819
Photodiodes 824
Photoconductive Cells 827
IR Emitters 829
Liquid-Crystal Displays 831
Solar Cells 833
Thermistors 837
pnpn AND OTHER DEVICES
842
Introduction 842
Silicon-Controlled Rectifier 842
Basic Silicon-Controlled Rectifier Operation 842
SCR Characteristics and Ratings 845
SCR Construction and Terminal Identification 847
SCR Applications 848
Silicon-Controlled Switch 852
Gate Turn-Off Switch 854
Light-Activated SCR 855
Shockley Diode 858
DIAC 858
TRIAC 860
Unijunction Transistor 861
Phototransistors 871
Opto-Isolators 873
Programmable Unijunction Transistor 875
Contents
xi
22
22.1
22.2
22.3
22.4
22.5
22.6
22.7
22.8
22.9
xii
Contents
OSCILLOSCOPE AND OTHER
MEASURING INSTRUMENTS
884
Introduction 884
Cathode Ray Tube—Theory and Construction 884
Cathode Ray Oscilloscope Operation 885
Voltage Sweep Operation 886
Synchronization and Triggering 889
Multitrace Operation 893
Measurement Using Calibrated CRO Scales 893
Special CRO Features 898
Signal Generators 899
APPENDIX A: HYBRID PARAMETERS—
CONVERSION EQUATIONS
(EXACT AND APPROXIMATE)
902
APPENDIX B: RIPPLE FACTOR AND
VOLTAGE CALCULATIONS
904
APPENDIX C: CHARTS AND TABLES
911
APPENDIX D: SOLUTIONS TO SELECTED
ODD-NUMBERED PROBLEMS
913
INDEX
919
Acknowledgments
Our sincerest appreciation must be extended to the instructors who have used the text
and sent in comments, corrections, and suggestions. We also want to thank Rex Davidson, Production Editor at Prentice Hall, for keeping together the many detailed aspects of production. Our sincerest thanks to Dave Garza, Senior Editor, and Linda
Ludewig, Editor, at Prentice Hall for their editorial support of the Seventh Edition of
this text.
We wish to thank those individuals who have shared their suggestions and evaluations of this text throughout its many editions. The comments from these individuals have enabled us to present Electronic Devices and Circuit Theory in this Seventh
Edition:
Ernest Lee Abbott
Phillip D. Anderson
Al Anthony
A. Duane Bailey
Joe Baker
Jerrold Barrosse
Ambrose Barry
Arthur Birch
Scott Bisland
Edward Bloch
Gary C. Bocksch
Jeffrey Bowe
Alfred D. Buerosse
Lila Caggiano
Mauro J. Caputi
Robert Casiano
Alan H. Czarapata
Mohammad Dabbas
John Darlington
Lucius B. Day
Mike Durren
Dr. Stephen Evanson
George Fredericks
F. D. Fuller
Napa College, Napa, CA
Muskegon Community College, Muskegon, MI
EG&G VACTEC Inc.
Southern Alberta Institute of Technology, Calgary, Alberta, CANADA
University of Southern California, Los Angeles, CA
Penn State–Ogontz
University of North Carolina–Charlotte
Hartford State Technical College, Hartford, CT
SEMATECH, Austin, TX
The Perkin-Elmer Corporation
Charles S. Mott Community College, Flint, MI
Bunker Hill Community College, Charlestown, MA
Waukesha County Technical College, Pewaukee, WI
MicroSim Corporation
Hofstra University
International Rectifier Corporation
Montgomery College, Rockville, MD
ITT Technical Institute
Humber College, Ontario, CANADA
Metropolitan State College, Denver, CO
Indiana Vocational Technical College, South Bend, IN
Bradford University, UK
Northeast State Technical Community College, Blountville, TN
Humber College, Ontario, CANADA
xvii
Phil Golden
Joseph Grabinski
Thomas K. Grady
William Hill
Albert L. Ickstadt
Jeng-Nan Juang
Karen Karger
Kenneth E. Kent
Donald E. King
Charles Lewis
Donna Liverman
William Mack
Robert Martin
George T. Mason
William Maxwell
Abraham Michelen
John MacDougall
Donald E. McMillan
Thomas E. Newman
Byron Paul
Dr. Robert Payne
Dr. Robert A. Powell
E. F. Rockafellow
Saeed A. Shaikh
Dr. Noel Shammas
Ken Simpson
Eric Sung
Donald P. Szymanski
Parker M. Tabor
Peter Tampas
Chuck Tinney
Katherine L. Usik
Domingo Uy
Richard J. Walters
Larry J. Wheeler
Julian Wilson
Syd R. Wilson
Jean Younes
Charles E. Yunghans
Ulrich E. Zeisler
xviii
Acknowledgments
DeVry Institute of Technology, Irving, TX
Hartford State Technical College, Hartfold, CT
Western Washington University, Bellingham, WA
ITT Technical Institute
San Diego Mesa College, San Diego, CA
Mercer University, Macon, GA
Tektronix Inc.
DeKalb Technical Institute, Clarkston, GA
ITT Technical Institute, Youngstown, OH
APPLIED MATERIALS, INC.
Texas Instruments Inc.
Harrisburg Area Community College
Northern Virginia Community College
Indiana Vocational Technical College, South Bend, IN
Nashville State Technical Institute
Hudson Valley Community College
University of Western Ontario, London, Ontario,
CANADA
Southwest State University, Marshall, MN
L. H. Bates Vocational-Technical Institute, Tacoma, WA
Bismarck State College
University of Glamorgan, Wales, UK
Oakland Community College
Southern-Alberta Institute of Technology, Calgary,
Alberta, CANADA
Miami-Dade Community College, Miami, FL
School of Engineering, Beaconside, UK
Stark State College of Technology
Computronics Technology Inc.
Owens Technical College, Toledo, OH
Greenville Technical College, Greenville, SC
Michigan Technological University, Houghton, MI
University of Utah
Mohawk College of Applied Art & Technology,
Hamilton, Ontario, CANADA
Hampton University, Hampton, VA
DeVry Technical Institute, Woodbridge, NJ
PSE&G Nuclear
Southern College of Technology, Marietta, GA
Motorola Inc.
ITT Technical Institute, Troy, MI
Western Washington University, Bellingham, WA
Salt Lake Community College, Salt Lake City, UT
p n
CHAPTER
Semiconductor
Diodes
1
1.1 INTRODUCTION
It is now some 50 years since the first transistor was introduced on December 23,
1947. For those of us who experienced the change from glass envelope tubes to the
solid-state era, it still seems like a few short years ago. The first edition of this text
contained heavy coverage of tubes, with succeeding editions involving the important
decision of how much coverage should be dedicated to tubes and how much to semiconductor devices. It no longer seems valid to mention tubes at all or to compare the
advantages of one over the other—we are firmly in the solid-state era.
The miniaturization that has resulted leaves us to wonder about its limits. Complete systems now appear on wafers thousands of times smaller than the single element of earlier networks. New designs and systems surface weekly. The engineer becomes more and more limited in his or her knowledge of the broad range of advances—
it is difficult enough simply to stay abreast of the changes in one area of research or
development. We have also reached a point at which the primary purpose of the container is simply to provide some means of handling the device or system and to provide a mechanism for attachment to the remainder of the network. Miniaturization
appears to be limited by three factors (each of which will be addressed in this text):
the quality of the semiconductor material itself, the network design technique, and
the limits of the manufacturing and processing equipment.
1.2 IDEAL DIODE
The first electronic device to be introduced is called the diode. It is the simplest of
semiconductor devices but plays a very vital role in electronic systems, having characteristics that closely match those of a simple switch. It will appear in a range of applications, extending from the simple to the very complex. In addition to the details
of its construction and characteristics, the very important data and graphs to be found
on specification sheets will also be covered to ensure an understanding of the terminology employed and to demonstrate the wealth of information typically available
from manufacturers.
The term ideal will be used frequently in this text as new devices are introduced.
It refers to any device or system that has ideal characteristics—perfect in every way.
It provides a basis for comparison, and it reveals where improvements can still be
made. The ideal diode is a two-terminal device having the symbol and characteristics shown in Figs. 1.1a and b, respectively.
Figure 1.1 Ideal diode: (a)
symbol; (b) characteristics.
1
p n
Ideally, a diode will conduct current in the direction defined by the arrow in the
symbol and act like an open circuit to any attempt to establish current in the opposite direction. In essence:
The characteristics of an ideal diode are those of a switch that can conduct
current in only one direction.
In the description of the elements to follow, it is critical that the various letter
symbols, voltage polarities, and current directions be defined. If the polarity of the
applied voltage is consistent with that shown in Fig. 1.1a, the portion of the characteristics to be considered in Fig. 1.1b is to the right of the vertical axis. If a reverse
voltage is applied, the characteristics to the left are pertinent. If the current through
the diode has the direction indicated in Fig. 1.1a, the portion of the characteristics to
be considered is above the horizontal axis, while a reversal in direction would require
the use of the characteristics below the axis. For the majority of the device characteristics that appear in this book, the ordinate (or “y” axis) will be the current axis,
while the abscissa (or “x” axis) will be the voltage axis.
One of the important parameters for the diode is the resistance at the point or region of operation. If we consider the conduction region defined by the direction of ID
and polarity of VD in Fig. 1.1a (upper-right quadrant of Fig. 1.1b), we will find that
the value of the forward resistance, RF, as defined by Ohm’s law is
VF
0V
RF 0 IF
2, 3, mA, . . . , or any positive value
(short circuit)
where VF is the forward voltage across the diode and IF is the forward current through
the diode.
The ideal diode, therefore, is a short circuit for the region of conduction.
Consider the region of negatively applied potential (third quadrant) of Fig. 1.1b,
5, 20, or any reverse-bias potential
VR
RR IR
0 mA
(open-circuit)
where VR is reverse voltage across the diode and IR is reverse current in the diode.
The ideal diode, therefore, is an open circuit in the region of nonconduction.
In review, the conditions depicted in Fig. 1.2 are applicable.
+
VD
–
Short circuit
ID
I D (limited by circuit)
(a)
0
–
VD
+
VD
Open circuit
ID = 0
(b)
Figure 1.2 (a) Conduction and (b) nonconduction states of the ideal diode as
determined by the applied bias.
In general, it is relatively simple to determine whether a diode is in the region of
conduction or nonconduction simply by noting the direction of the current ID established by an applied voltage. For conventional flow (opposite to that of electron flow),
if the resultant diode current has the same direction as the arrowhead of the diode
symbol, the diode is operating in the conducting region as depicted in Fig. 1.3a. If
2
Chapter 1
Semiconductor Diodes
p n
the resulting current has the opposite direction, as shown in Fig. 1.3b, the opencircuit equivalent is appropriate.
ID
ID
(a)
ID = 0
ID
Figure 1.3 (a) Conduction
and (b) nonconduction states of
the ideal diode as determined by
the direction of conventional
current established by the
network.
(b)
As indicated earlier, the primary purpose of this section is to introduce the characteristics of an ideal device for comparison with the characteristics of the commercial variety. As we progress through the next few sections, keep the following questions in mind:
How close will the forward or “on” resistance of a practical diode compare
with the desired 0- level?
Is the reverse-bias resistance sufficiently large to permit an open-circuit approximation?
1.3 SEMICONDUCTOR MATERIALS
The label semiconductor itself provides a hint as to its characteristics. The prefix semiis normally applied to a range of levels midway between two limits.
The term conductor is applied to any material that will support a generous
flow of charge when a voltage source of limited magnitude is applied across
its terminals.
An insulator is a material that offers a very low level of conductivity under
pressure from an applied voltage source.
A semiconductor, therefore, is a material that has a conductivity level somewhere between the extremes of an insulator and a conductor.
Inversely related to the conductivity of a material is its resistance to the flow of
charge, or current. That is, the higher the conductivity level, the lower the resistance
level. In tables, the term resistivity (, Greek letter rho) is often used when comparing the resistance levels of materials. In metric units, the resistivity of a material is
measured in -cm or -m. The units of -cm are derived from the substitution of
the units for each quantity of Fig. 1.4 into the following equation (derived from the
basic resistance equation R l/A):
RA
()(cm2)
⇒ -cm
l
cm
(1.1)
In fact, if the area of Fig. 1.4 is 1 cm2 and the length 1 cm, the magnitude of the
resistance of the cube of Fig. 1.4 is equal to the magnitude of the resistivity of the
material as demonstrated below:
Figure 1.4 Defining the metric
units of resistivity.
l
(1 cm)
ohms
R A
(1 cm2)
This fact will be helpful to remember as we compare resistivity levels in the discussions to follow.
In Table 1.1, typical resistivity values are provided for three broad categories of
materials. Although you may be familiar with the electrical properties of copper and
1.3 Semiconductor Materials
3
p n
TABLE 1.1 Typical Resistivity Values
Figure 1.5 Ge and Si
single-crystal structure.
4
Conductor
Semiconductor
Insulator
106 -cm
(copper)
50 -cm (germanium)
50 103 -cm (silicon)
1012 -cm
(mica)
mica from your past studies, the characteristics of the semiconductor materials of germanium (Ge) and silicon (Si) may be relatively new. As you will find in the chapters
to follow, they are certainly not the only two semiconductor materials. They are, however, the two materials that have received the broadest range of interest in the development of semiconductor devices. In recent years the shift has been steadily toward
silicon and away from germanium, but germanium is still in modest production.
Note in Table 1.1 the extreme range between the conductor and insulating materials for the 1-cm length (1-cm2 area) of the material. Eighteen places separate the
placement of the decimal point for one number from the other. Ge and Si have received the attention they have for a number of reasons. One very important consideration is the fact that they can be manufactured to a very high purity level. In fact,
recent advances have reduced impurity levels in the pure material to 1 part in 10 billion (110,000,000,000). One might ask if these low impurity levels are really necessary. They certainly are if you consider that the addition of one part impurity (of
the proper type) per million in a wafer of silicon material can change that material
from a relatively poor conductor to a good conductor of electricity. We are obviously
dealing with a whole new spectrum of comparison levels when we deal with the semiconductor medium. The ability to change the characteristics of the material significantly through this process, known as “doping,” is yet another reason why Ge and Si
have received such wide attention. Further reasons include the fact that their characteristics can be altered significantly through the application of heat or light—an important consideration in the development of heat- and light-sensitive devices.
Some of the unique qualities of Ge and Si noted above are due to their atomic
structure. The atoms of both materials form a very definite pattern that is periodic in
nature (i.e., continually repeats itself). One complete pattern is called a crystal and
the periodic arrangement of the atoms a lattice. For Ge and Si the crystal has the
three-dimensional diamond structure of Fig. 1.5. Any material composed solely of repeating crystal structures of the same kind is called a single-crystal structure. For
semiconductor materials of practical application in the electronics field, this singlecrystal feature exists, and, in addition, the periodicity of the structure does not change
significantly with the addition of impurities in the doping process.
Let us now examine the structure of the atom itself and note how it might affect
the electrical characteristics of the material. As you are aware, the atom is composed
of three basic particles: the electron, the proton, and the neutron. In the atomic lattice, the neutrons and protons form the nucleus, while the electrons revolve around
the nucleus in a fixed orbit. The Bohr models of the two most commonly used semiconductors, germanium and silicon, are shown in Fig. 1.6.
As indicated by Fig. 1.6a, the germanium atom has 32 orbiting electrons, while
silicon has 14 orbiting electrons. In each case, there are 4 electrons in the outermost
(valence) shell. The potential (ionization potential) required to remove any one of
these 4 valence electrons is lower than that required for any other electron in the structure. In a pure germanium or silicon crystal these 4 valence electrons are bonded to
4 adjoining atoms, as shown in Fig. 1.7 for silicon. Both Ge and Si are referred to as
tetravalent atoms because they each have four valence electrons.
A bonding of atoms, strengthened by the sharing of electrons, is called covalent bonding.
Chapter 1
Semiconductor Diodes
p n
Figure 1.6 Atomic structure: (a) germanium;
(b) silicon.
Figure 1.7
atom.
Covalent bonding of the silicon
Although the covalent bond will result in a stronger bond between the valence
electrons and their parent atom, it is still possible for the valence electrons to absorb
sufficient kinetic energy from natural causes to break the covalent bond and assume
the “free” state. The term free reveals that their motion is quite sensitive to applied
electric fields such as established by voltage sources or any difference in potential.
These natural causes include effects such as light energy in the form of photons and
thermal energy from the surrounding medium. At room temperature there are approximately 1.5 1010 free carriers in a cubic centimeter of intrinsic silicon material.
Intrinsic materials are those semiconductors that have been carefully refined
to reduce the impurities to a very low level—essentially as pure as can be
made available through modern technology.
The free electrons in the material due only to natural causes are referred to as
intrinsic carriers. At the same temperature, intrinsic germanium material will have
approximately 2.5 1013 free carriers per cubic centimeter. The ratio of the number of carriers in germanium to that of silicon is greater than 103 and would indicate that germanium is a better conductor at room temperature. This may be true,
but both are still considered poor conductors in the intrinsic state. Note in Table 1.1
that the resistivity also differs by a ratio of about 10001, with silicon having the
larger value. This should be the case, of course, since resistivity and conductivity are
inversely related.
An increase in temperature of a semiconductor can result in a substantial increase in the number of free electrons in the material.
As the temperature rises from absolute zero (0 K), an increasing number of valence electrons absorb sufficient thermal energy to break the covalent bond and contribute to the number of free carriers as described above. This increased number of
carriers will increase the conductivity index and result in a lower resistance level.
Semiconductor materials such as Ge and Si that show a reduction in resistance with increase in temperature are said to have a negative temperature
coefficient.
You will probably recall that the resistance of most conductors will increase with
temperature. This is due to the fact that the numbers of carriers in a conductor will
1.3 Semiconductor Materials
5
p n
not increase significantly with temperature, but their vibration pattern about a relatively fixed location will make it increasingly difficult for electrons to pass through.
An increase in temperature therefore results in an increased resistance level and a positive temperature coefficient.
1.4 ENERGY LEVELS
In the isolated atomic structure there are discrete (individual) energy levels associated
with each orbiting electron, as shown in Fig. 1.8a. Each material will, in fact, have
its own set of permissible energy levels for the electrons in its atomic structure.
The more distant the electron from the nucleus, the higher the energy state,
and any electron that has left its parent atom has a higher energy state than
any electron in the atomic structure.
Energy
Valance Level (outermost shell)
Energy gap
Second Level (next inner shell)
Energy gap
Third Level (etc.)
etc.
Nucleus
(a)
Energy
Conduction band
Electrons
"free" to
establish
conduction
Energy
Conduction band
Eg
E g > 5 eV
Valence band
Figure 1.8 Energy levels: (a)
discrete levels in isolated atomic
structures; (b) conduction and
valence bands of an insulator,
semiconductor, and conductor.
Energy
Valence
electrons
bound to
the atomic
stucture
Insulator
The bands
overlap
Conduction band
Valence band
Valence band
E g = 1.1 eV (Si)
E g = 0.67 eV (Ge)
E g = 1.41 eV (GaAs)
Semiconductor
Conductor
(b)
Between the discrete energy levels are gaps in which no electrons in the isolated
atomic structure can appear. As the atoms of a material are brought closer together to
form the crystal lattice structure, there is an interaction between atoms that will result in the electrons in a particular orbit of one atom having slightly different energy
levels from electrons in the same orbit of an adjoining atom. The net result is an expansion of the discrete levels of possible energy states for the valence electrons to
that of bands as shown in Fig. 1.8b. Note that there are boundary levels and maximum energy states in which any electron in the atomic lattice can find itself, and there
remains a forbidden region between the valence band and the ionization level. Recall
6
Chapter 1
Semiconductor Diodes
p n
that ionization is the mechanism whereby an electron can absorb sufficient energy to
break away from the atomic structure and enter the conduction band. You will note
that the energy associated with each electron is measured in electron volts (eV). The
unit of measure is appropriate, since
W QV
eV
(1.2)
as derived from the defining equation for voltage V W/Q. The charge Q is the charge
associated with a single electron.
Substituting the charge of an electron and a potential difference of 1 volt into Eq.
(1.2) will result in an energy level referred to as one electron volt. Since energy is
also measured in joules and the charge of one electron 1.6 1019 coulomb,
W QV (1.6 1019 C)(1 V)
and
1 eV 1.6 1019 J
(1.3)
At 0 K or absolute zero (273.15°C), all the valence electrons of semiconductor
materials find themselves locked in their outermost shell of the atom with energy
levels associated with the valence band of Fig. 1.8b. However, at room temperature
(300 K, 25°C) a large number of valence electrons have acquired sufficient energy to
leave the valence band, cross the energy gap defined by Eg in Fig. 1.8b and enter the
conduction band. For silicon Eg is 1.1 eV, for germanium 0.67 eV, and for gallium
arsenide 1.41 eV. The obviously lower Eg for germanium accounts for the increased
number of carriers in that material as compared to silicon at room temperature. Note
for the insulator that the energy gap is typically 5 eV or more, which severely limits
the number of electrons that can enter the conduction band at room temperature. The
conductor has electrons in the conduction band even at 0 K. Quite obviously, therefore, at room temperature there are more than enough free carriers to sustain a heavy
flow of charge, or current.
We will find in Section 1.5 that if certain impurities are added to the intrinsic
semiconductor materials, energy states in the forbidden bands will occur which will
cause a net reduction in Eg for both semiconductor materials—consequently, increased
carrier density in the conduction band at room temperature!
1.5 EXTRINSIC MATERIALS—
n- AND p-TYPE
The characteristics of semiconductor materials can be altered significantly by the addition of certain impurity atoms into the relatively pure semiconductor material. These
impurities, although only added to perhaps 1 part in 10 million, can alter the band
structure sufficiently to totally change the electrical properties of the material.
A semiconductor material that has been subjected to the doping process is
called an extrinsic material.
There are two extrinsic materials of immeasurable importance to semiconductor
device fabrication: n-type and p-type. Each will be described in some detail in the
following paragraphs.
n-Type Material
Both the n- and p-type materials are formed by adding a predetermined number of
impurity atoms into a germanium or silicon base. The n-type is created by introducing those impurity elements that have five valence electrons (pentavalent), such as antimony, arsenic, and phosphorus. The effect of such impurity elements is indicated in
1.5 Extrinsic Materials—n- and p-Type
7
p n
–
–
Si
–
–
–
Si
–
Si
–
–
Si
–
–
–
Si
–
–
Si
–
–
– –
– Sb –
–
–
–
–
–
–
–
–
–
Fifth valence
electron
of antimony
–
–
Si
–
–
Antimony (Sb)
impurity
–
–
–
Si
–
–
Figure 1.9 Antimony impurity
in n-type material.
Fig. 1.9 (using antimony as the impurity in a silicon base). Note that the four covalent bonds are still present. There is, however, an additional fifth electron due to the
impurity atom, which is unassociated with any particular covalent bond. This remaining electron, loosely bound to its parent (antimony) atom, is relatively free to
move within the newly formed n-type material. Since the inserted impurity atom has
donated a relatively “free” electron to the structure:
Diffused impurities with five valence electrons are called donor atoms.
It is important to realize that even though a large number of “free” carriers have
been established in the n-type material, it is still electrically neutral since ideally the
number of positively charged protons in the nuclei is still equal to the number of
“free” and orbiting negatively charged electrons in the structure.
The effect of this doping process on the relative conductivity can best be described
through the use of the energy-band diagram of Fig. 1.10. Note that a discrete energy
level (called the donor level) appears in the forbidden band with an Eg significantly
less than that of the intrinsic material. Those “free” electrons due to the added impurity sit at this energy level and have less difficulty absorbing a sufficient measure
of thermal energy to move into the conduction band at room temperature. The result
is that at room temperature, there are a large number of carriers (electrons) in the
conduction level and the conductivity of the material increases significantly. At room
temperature in an intrinsic Si material there is about one free electron for every 1012
atoms (1 to 109 for Ge). If our dosage level were 1 in 10 million (107), the ratio
(1012/107 105) would indicate that the carrier concentration has increased by a ratio of 100,0001.
Energy
Conduction band
E g = 0.05 eV (Si), 0.01 eV (Ge)
Donor energy level
E g as before
Valence band
Figure 1.10 Effect of donor impurities on the energy band
structure.
8
Chapter 1
Semiconductor Diodes
p n
p-Type Material
The p-type material is formed by doping a pure germanium or silicon crystal with
impurity atoms having three valence electrons. The elements most frequently used for
this purpose are boron, gallium, and indium. The effect of one of these elements,
boron, on a base of silicon is indicated in Fig. 1.11.
Figure 1.11 Boron impurity in
p-type material.
Note that there is now an insufficient number of electrons to complete the covalent bonds of the newly formed lattice. The resulting vacancy is called a hole and is
represented by a small circle or positive sign due to the absence of a negative charge.
Since the resulting vacancy will readily accept a “free” electron:
The diffused impurities with three valence electrons are called acceptor atoms.
The resulting p-type material is electrically neutral, for the same reasons described
for the n-type material.
Electron versus Hole Flow
The effect of the hole on conduction is shown in Fig. 1.12. If a valence electron acquires sufficient kinetic energy to break its covalent bond and fills the void created
by a hole, then a vacancy, or hole, will be created in the covalent bond that released
the electron. There is, therefore, a transfer of holes to the left and electrons to the
right, as shown in Fig. 1.12. The direction to be used in this text is that of conventional flow, which is indicated by the direction of hole flow.
Figure 1.12 Electron versus
hole flow.
1.5 Extrinsic Materials—n- and p-Type
9
p n
Majority and Minority Carriers
In the intrinsic state, the number of free electrons in Ge or Si is due only to those few
electrons in the valence band that have acquired sufficient energy from thermal or
light sources to break the covalent bond or to the few impurities that could not be removed. The vacancies left behind in the covalent bonding structure represent our very
limited supply of holes. In an n-type material, the number of holes has not changed
significantly from this intrinsic level. The net result, therefore, is that the number of
electrons far outweighs the number of holes. For this reason:
In an n-type material (Fig. 1.13a) the electron is called the majority carrier
and the hole the minority carrier.
For the p-type material the number of holes far outweighs the number of electrons, as shown in Fig. 1.13b. Therefore:
In a p-type material the hole is the majority carrier and the electron is the
minority carrier.
When the fifth electron of a donor atom leaves the parent atom, the atom remaining
acquires a net positive charge: hence the positive sign in the donor-ion representation.
For similar reasons, the negative sign appears in the acceptor ion.
The n- and p-type materials represent the basic building blocks of semiconductor
devices. We will find in the next section that the “joining” of a single n-type material with a p-type material will result in a semiconductor element of considerable importance in electronic systems.
Acceptor ions
Donor ions
+ ––
– +
–
+
+ –
– +
+ – + –
+ + –
+
– –
+
–
+ +
–
– +
Majority
carriers
Minority
carrier
Majority
carriers
+
–
+ –
– + +–
– + + – –+ +
+ –
+ + –
–
+
+
– + – + –
n-type
p-type
(a)
(b)
Minority
carrier
Figure 1.13 (a) n-type material; (b) p-type material.
1.6 SEMICONDUCTOR DIODE
In Section 1.5 both the n- and p-type materials were introduced. The semiconductor
diode is formed by simply bringing these materials together (constructed from the
same base—Ge or Si), as shown in Fig. 1.14, using techniques to be described in
Chapter 20. At the instant the two materials are “joined” the electrons and holes in
the region of the junction will combine, resulting in a lack of carriers in the region
near the junction.
This region of uncovered positive and negative ions is called the depletion region due to the depletion of carriers in this region.
Since the diode is a two-terminal device, the application of a voltage across its
terminals leaves three possibilities: no bias (VD 0 V), forward bias (VD 0 V), and
reverse bias (VD 0 V). Each is a condition that will result in a response that the
user must clearly understand if the device is to be applied effectively.
10
Chapter 1
Semiconductor Diodes
p n
Figure 1.14 p-n junction with
no external bias.
No Applied Bias (VD 0 V)
Under no-bias (no applied voltage) conditions, any minority carriers (holes) in the
n-type material that find themselves within the depletion region will pass directly into
the p-type material. The closer the minority carrier is to the junction, the greater the
attraction for the layer of negative ions and the less the opposition of the positive ions
in the depletion region of the n-type material. For the purposes of future discussions
we shall assume that all the minority carriers of the n-type material that find themselves in the depletion region due to their random motion will pass directly into the
p-type material. Similar discussion can be applied to the minority carriers (electrons)
of the p-type material. This carrier flow has been indicated in Fig. 1.14 for the minority carriers of each material.
The majority carriers (electrons) of the n-type material must overcome the attractive forces of the layer of positive ions in the n-type material and the shield of
negative ions in the p-type material to migrate into the area beyond the depletion region of the p-type material. However, the number of majority carriers is so large in
the n-type material that there will invariably be a small number of majority carriers
with sufficient kinetic energy to pass through the depletion region into the p-type material. Again, the same type of discussion can be applied to the majority carriers (holes)
of the p-type material. The resulting flow due to the majority carriers is also shown
in Fig. 1.14.
A close examination of Fig. 1.14 will reveal that the relative magnitudes of the
flow vectors are such that the net flow in either direction is zero. This cancellation of
vectors has been indicated by crossed lines. The length of the vector representing hole
flow has been drawn longer than that for electron flow to demonstrate that the magnitude of each need not be the same for cancellation and that the doping levels for
each material may result in an unequal carrier flow of holes and electrons. In summary, therefore:
In the absence of an applied bias voltage, the net flow of charge in any one
direction for a semiconductor diode is zero.
1.6 Semiconductor Diode
11
p n
The symbol for a diode is repeated in Fig. 1.15 with the associated n- and p-type
regions. Note that the arrow is associated with the p-type component and the bar with
the n-type region. As indicated, for VD 0 V, the current in any direction is 0 mA.
Reverse-Bias Condition (VD
Figure 1.15 No-bias conditions
for a semiconductor diode.
0 V)
If an external potential of V volts is applied across the p-n junction such that the positive terminal is connected to the n-type material and the negative terminal is connected to the p-type material as shown in Fig. 1.16, the number of uncovered positive ions in the depletion region of the n-type material will increase due to the large
number of “free” electrons drawn to the positive potential of the applied voltage. For
similar reasons, the number of uncovered negative ions will increase in the p-type
material. The net effect, therefore, is a widening of the depletion region. This widening of the depletion region will establish too great a barrier for the majority carriers to
overcome, effectively reducing the majority carrier flow to zero as shown in Fig. 1.16.
Figure 1.16 Reverse-biased
p-n junction.
The number of minority carriers, however, that find themselves entering the depletion region will not change, resulting in minority-carrier flow vectors of the same
magnitude indicated in Fig. 1.14 with no applied voltage.
The current that exists under reverse-bias conditions is called the reverse saturation current and is represented by Is.
Figure 1.17 Reverse-bias
conditions for a semiconductor
diode.
The reverse saturation current is seldom more than a few microamperes except for
high-power devices. In fact, in recent years its level is typically in the nanoampere
range for silicon devices and in the low-microampere range for germanium. The term
saturation comes from the fact that it reaches its maximum level quickly and does not
change significantly with increase in the reverse-bias potential, as shown on the diode
characteristics of Fig. 1.19 for VD 0 V. The reverse-biased conditions are depicted
in Fig. 1.17 for the diode symbol and p-n junction. Note, in particular, that the direction of Is is against the arrow of the symbol. Note also that the negative potential is
connected to the p-type material and the positive potential to the n-type material—the
difference in underlined letters for each region revealing a reverse-bias condition.
Forward-Bias Condition (VD
0 V)
A forward-bias or “on” condition is established by applying the positive potential to
the p-type material and the negative potential to the n-type material as shown in Fig.
1.18. For future reference, therefore:
A semiconductor diode is forward-biased when the association p-type and positive and n-type and negative has been established.
12
Chapter 1
Semiconductor Diodes
p n
Figure 1.18 Forward-biased p-n
junction.
The application of a forward-bias potential VD will “pressure” electrons in the
n-type material and holes in the p-type material to recombine with the ions near the
boundary and reduce the width of the depletion region as shown in Fig. 1.18. The resulting minority-carrier flow of electrons from the p-type material to the n-type material (and of holes from the n-type material to the p-type material) has not changed
in magnitude (since the conduction level is controlled primarily by the limited number of impurities in the material), but the reduction in the width of the depletion region has resulted in a heavy majority flow across the junction. An electron of the
n-type material now “sees” a reduced barrier at the junction due to the reduced depletion region and a strong attraction for the positive potential applied to the p-type
material. As the applied bias increases in magnitude the depletion region will continue to decrease in width until a flood of electrons can pass through the junction, reID (mA)
20
19
Eq. (1.4)
18
Actual commercially
available unit
17
16
15
14
13
12
Defined polarity and
direction for graph
VD
11
10
+
9
–
ID
8
7
Forward-bias region
(V
VD > 0 V, II D > 0 mA)
6
5
4
3
2
Is
–40
–30
–20
1
–10
Reverse-bias region
(VD < 0 V, ID = –Is )
0
0.3
– 0.1 µ
uA
– 0.2 µ
uA
– 0.3 µ
uA
0.5
0.7
1
V D (V)
No-bias
(VD = 0 V, ID = 0 mA)
– 0.4 µ
uA
Figure 1.19 Silicon semiconductor
diode characteristics.
1.6 Semiconductor Diode
13
p n
sulting in an exponential rise in current as shown in the forward-bias region of the
characteristics of Fig. 1.19. Note that the vertical scale of Fig. 1.19 is measured in
milliamperes (although some semiconductor diodes will have a vertical scale measured in amperes) and the horizontal scale in the forward-bias region has a maximum
of 1 V. Typically, therefore, the voltage across a forward-biased diode will be less
than 1 V. Note also, how quickly the current rises beyond the knee of the curve.
It can be demonstrated through the use of solid-state physics that the general characteristics of a semiconductor diode can be defined by the following equation for the
forward- and reverse-bias regions:
ID Is(ekVD/TK 1)
where
(1.4)
Is reverse saturation current
k 11,600/ with 1 for Ge and 2 for Si for relatively low levels
of diode current (at or below the knee of the curve) and 1 for Ge
and Si for higher levels of diode current (in the rapidly increasing section of the curve)
TK TC 273°
A plot of Eq. (1.4) is provided in Fig. 1.19. If we expand Eq. (1.4) into the following form, the contributing component for each region of Fig. 1.19 can easily be
described:
ID IsekVD/TK Is
Figure 1.20 Plot of e x.
For positive values of VD the first term of the equation above will grow very
quickly and overpower the effect of the second term. The result is that for positive
values of VD, ID will be positive and grow as the function y ex appearing in Fig.
1.20. At VD 0 V, Eq. (1.4) becomes ID Is(e0 1) Is(1 1) 0 mA as appearing in Fig. 1.19. For negative values of VD the first term will quickly drop off below Is, resulting in ID Is, which is simply the horizontal line of Fig. 1.19. The
break in the characteristics at VD 0 V is simply due to the dramatic change in scale
from mA to A.
Note in Fig. 1.19 that the commercially available unit has characteristics that are
shifted to the right by a few tenths of a volt. This is due to the internal “body” resistance and external “contact” resistance of a diode. Each contributes to an additional
voltage at the same current level as determined by Ohm’s law (V IR). In time, as
production methods improve, this difference will decrease and the actual characteristics approach those of Eq. (1.4).
It is important to note the change in scale for the vertical and horizontal axes. For
positive values of ID the scale is in milliamperes and the current scale below the axis
is in microamperes (or possibly nanoamperes). For VD the scale for positive values is
in tenths of volts and for negative values the scale is in tens of volts.
Initially, Eq. (1.4) does appear somewhat complex and may develop an unwarranted fear that it will be applied for all the diode applications to follow. Fortunately,
however, a number of approximations will be made in a later section that will negate
the need to apply Eq. (1.4) and provide a solution with a minimum of mathematical
difficulty.
Before leaving the subject of the forward-bias state the conditions for conduction
(the “on” state) are repeated in Fig. 1.21 with the required biasing polarities and the
resulting direction of majority-carrier flow. Note in particular how the direction of
conduction matches the arrow in the symbol (as revealed for the ideal diode).
Zener Region
Figure 1.21 Forward-bias
conditions for a semiconductor
diode.
Even though the scale of Fig. 1.19 is in tens of volts in the negative region, there is
a point where the application of too negative a voltage will result in a sharp change
14
Chapter 1
Semiconductor Diodes
p n
Figure 1.22 Zener region.
in the characteristics, as shown in Fig. 1.22. The current increases at a very rapid rate
in a direction opposite to that of the positive voltage region. The reverse-bias potential that results in this dramatic change in characteristics is called the Zener potential
and is given the symbol VZ.
As the voltage across the diode increases in the reverse-bias region, the velocity
of the minority carriers responsible for the reverse saturation current Is will also increase. Eventually, their velocity and associated kinetic energy (WK 12mv2) will be
sufficient to release additional carriers through collisions with otherwise stable atomic
structures. That is, an ionization process will result whereby valence electrons absorb
sufficient energy to leave the parent atom. These additional carriers can then aid the
ionization process to the point where a high avalanche current is established and the
avalanche breakdown region determined.
The avalanche region (VZ) can be brought closer to the vertical axis by increasing
the doping levels in the p- and n-type materials. However, as VZ decreases to very low
levels, such as 5 V, another mechanism, called Zener breakdown, will contribute to
the sharp change in the characteristic. It occurs because there is a strong electric field
in the region of the junction that can disrupt the bonding forces within the atom and
“generate” carriers. Although the Zener breakdown mechanism is a significant contributor only at lower levels of VZ, this sharp change in the characteristic at any level is
called the Zener region and diodes employing this unique portion of the characteristic
of a p-n junction are called Zener diodes. They are described in detail in Section 1.14.
The Zener region of the semiconductor diode described must be avoided if the response of a system is not to be completely altered by the sharp change in characteristics in this reverse-voltage region.
The maximum reverse-bias potential that can be applied before entering the
Zener region is called the peak inverse voltage (referred to simply as the PIV
rating) or the peak reverse voltage (denoted by PRV rating).
If an application requires a PIV rating greater than that of a single unit, a number of diodes of the same characteristics can be connected in series. Diodes are also
connected in parallel to increase the current-carrying capacity.
Silicon versus Germanium
Silicon diodes have, in general, higher PIV and current rating and wider temperature
ranges than germanium diodes. PIV ratings for silicon can be in the neighborhood of
1000 V, whereas the maximum value for germanium is closer to 400 V. Silicon can
be used for applications in which the temperature may rise to about 200°C (400°F),
whereas germanium has a much lower maximum rating (100°C). The disadvantage
of silicon, however, as compared to germanium, as indicated in Fig. 1.23, is the higher
1.6 Semiconductor Diode
15
p n
Figure 1.23 Comparison of Si
and Ge semiconductor diodes.
forward-bias voltage required to reach the region of upward swing. It is typically of
the order of magnitude of 0.7 V for commercially available silicon diodes and 0.3 V
for germanium diodes when rounded off to the nearest tenths. The increased offset
for silicon is due primarily to the factor in Eq. (1.4). This factor plays a part in determining the shape of the curve only at very low current levels. Once the curve starts
its vertical rise, the factor drops to 1 (the continuous value for germanium). This is
evidenced by the similarities in the curves once the offset potential is reached. The
potential at which this rise occurs is commonly referred to as the offset, threshold, or
firing potential. Frequently, the first letter of a term that describes a particular quantity is used in the notation for that quantity. However, to ensure a minimum of confusion with other terms, such as output voltage (Vo) and forward voltage (VF), the notation VT has been adopted for this book, from the word “threshold.”
In review:
VT 0.7 (Si)
VT 0.3 (Ge)
Obviously, the closer the upward swing is to the vertical axis, the more “ideal” the
device. However, the other characteristics of silicon as compared to germanium still
make it the choice in the majority of commercially available units.
Temperature Effects
Temperature can have a marked effect on the characteristics of a silicon semiconductor diode as witnessed by a typical silicon diode in Fig. 1.24. It has been found
experimentally that:
The reverse saturation current Is will just about double in magnitude for
every 10°C increase in temperature.
16
Chapter 1
Semiconductor Diodes
p n
Figure 1.24 Variation in diode
characteristics with temperature
change.
It is not uncommon for a germanium diode with an Is in the order of 1 or 2 A
at 25°C to have a leakage current of 100 A 0.1 mA at a temperature of 100°C.
Current levels of this magnitude in the reverse-bias region would certainly question
our desired open-circuit condition in the reverse-bias region. Typical values of Is for
silicon are much lower than that of germanium for similar power and current levels
as shown in Fig. 1.23. The result is that even at high temperatures the levels of Is for
silicon diodes do not reach the same high levels obtained for germanium—a very important reason that silicon devices enjoy a significantly higher level of development
and utilization in design. Fundamentally, the open-circuit equivalent in the reversebias region is better realized at any temperature with silicon than with germanium.
The increasing levels of Is with temperature account for the lower levels of threshold voltage, as shown in Fig. 1.24. Simply increase the level of Is in Eq. (1.4) and
note the earlier rise in diode current. Of course, the level of TK also will be increasing in the same equation, but the increasing level of Is will overpower the smaller percent change in TK. As the temperature increases the forward characteristics are actually becoming more “ideal,” but we will find when we review the specifications sheets
that temperatures beyond the normal operating range can have a very detrimental effect on the diode’s maximum power and current levels. In the reverse-bias region the
breakdown voltage is increasing with temperature, but note the undesirable increase
in reverse saturation current.
1.7 RESISTANCE LEVELS
As the operating point of a diode moves from one region to another the resistance of
the diode will also change due to the nonlinear shape of the characteristic curve. It
will be demonstrated in the next few paragraphs that the type of applied voltage or
signal will define the resistance level of interest. Three different levels will be introduced in this section that will appear again as we examine other devices. It is therefore paramount that their determination be clearly understood.
1.7
Resistance Levels
17
p n
DC or Static Resistance
The application of a dc voltage to a circuit containing a semiconductor diode will result in an operating point on the characteristic curve that will not change with time.
The resistance of the diode at the operating point can be found simply by finding the
corresponding levels of VD and ID as shown in Fig. 1.25 and applying the following
equation:
VD
RD ID
(1.5)
The dc resistance levels at the knee and below will be greater than the resistance
levels obtained for the vertical rise section of the characteristics. The resistance levels in the reverse-bias region will naturally be quite high. Since ohmmeters typically
employ a relatively constant-current source, the resistance determined will be at a preset current level (typically, a few milliamperes).
Figure 1.25 Determining the dc
resistance of a diode at a particular operating point.
In general, therefore, the lower the current through a diode the higher the dc
resistance level.
EXAMPLE 1.1
Determine the dc resistance levels for the diode of Fig. 1.26 at
(a) ID 2 mA
(b) ID 20 mA
(c) VD 10 V
Figure 1.26 Example 1.1
Solution
(a) At ID 2 mA, VD 0.5 V (from the curve) and
VD
0.5 V
RD 250 ID
2 mA
18
Chapter 1
Semiconductor Diodes
p n
(b) At ID 20 mA, VD 0.8 V (from the curve) and
VD
0.8 V
RD 40 ID
20 mA
(c) At VD 10 V, ID Is 1 A (from the curve) and
VD
10 V
RD 10 M
ID
1 A
clearly supporting some of the earlier comments regarding the dc resistance levels of
a diode.
AC or Dynamic Resistance
It is obvious from Eq. 1.5 and Example 1.1 that the dc resistance of a diode is independent of the shape of the characteristic in the region surrounding the point of interest. If a sinusoidal rather than dc input is applied, the situation will change completely.
The varying input will move the instantaneous operating point up and down a region
of the characteristics and thus defines a specific change in current and voltage as shown
in Fig. 1.27. With no applied varying signal, the point of operation would
be the Q-point appearing on Fig. 1.27 determined by the applied dc levels. The designation Q-point is derived from the word quiescent, which means “still or unvarying.”
Figure 1.27 Defining the
dynamic or ac resistance.
A straight line drawn tangent to the curve through the Q-point as shown in Fig.
1.28 will define a particular change in voltage and current that can be used to determine the ac or dynamic resistance for this region of the diode characteristics. An effort should be made to keep the change in voltage and current as small as possible
and equidistant to either side of the Q-point. In equation form,
Vd
rd Id
where signifies a finite change in the quantity.
(1.6)
The steeper the slope, the less the value of Vd for the same change in Id and the
less the resistance. The ac resistance in the vertical-rise region of the characteristic is
therefore quite small, while the ac resistance is much higher at low current levels.
In general, therefore, the lower the Q-point of operation (smaller current or
lower voltage) the higher the ac resistance.
1.7
Resistance Levels
Figure 1.28 Determining the ac
resistance at a Q-point.
19
p n
EXAMPLE 1.2
For the characteristics of Fig. 1.29:
(a) Determine the ac resistance at ID 2 mA.
(b) Determine the ac resistance at ID 25 mA.
(c) Compare the results of parts (a) and (b) to the dc resistances at each current level.
I D (mA)
30
∆ Id
25
20
∆Vd
15
10
5
4
2
∆ Id
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1
VD (V)
∆Vd
Figure 1.29 Example 1.2
Solution
(a) For ID 2 mA; the tangent line at ID 2 mA was drawn as shown in the figure
and a swing of 2 mA above and below the specified diode current was chosen.
At ID 4 mA, VD 0.76 V, and at ID 0 mA, VD 0.65 V. The resulting
changes in current and voltage are
Id 4 mA 0 mA 4 mA
Vd 0.76 V 0.65 V 0.11 V
and
and the ac resistance:
Vd
0.11 V
rd 27.5 Id
4 mA
(b) For ID 25 mA, the tangent line at ID 25 mA was drawn as shown on the figure and a swing of 5 mA above and below the specified diode current was chosen. At ID 30 mA, VD 0.8 V, and at ID 20 mA, VD 0.78 V. The resulting changes in current and voltage are
Id 30 mA 20 mA 10 mA
Vd 0.8 V 0.78 V 0.02 V
and
and the ac resistance is
20
Chapter 1
Vd
0.02 V
2 rd Id
10 mA
Semiconductor Diodes
p n
(c) For ID 2 mA, VD 0.7 V and
VD
0.7 V
RD 350 ID
2 mA
which far exceeds the rd of 27.5 .
For ID 25 mA, VD 0.79 V and
VD
0.79 V
RD 31.62 ID
25 mA
which far exceeds the rd of 2 .
We have found the dynamic resistance graphically, but there is a basic definition
in differential calculus which states:
The derivative of a function at a point is equal to the slope of the tangent line
drawn at that point.
Equation (1.6), as defined by Fig. 1.28, is, therefore, essentially finding the derivative of the function at the Q-point of operation. If we find the derivative of the general equation (1.4) for the semiconductor diode with respect to the applied forward
bias and then invert the result, we will have an equation for the dynamic or ac resistance in that region. That is, taking the derivative of Eq. (1.4) with respect to the applied bias will result in
d
d
(ID) [Is(ekVD /TK 1)]
dVD
dV
dID
k
(ID
dVD
TK
and
Is)
following a few basic maneuvers of differential calculus. In general, ID Is in the
vertical slope section of the characteristics and
dID
k
ID
dVD
TK
Substituting
obtain
1 for Ge and Si in the vertical-rise section of the characteristics, we
11,600
11,600
k 11,600
1
and at room temperature,
TK TC
so that
and
273° 25°
273° 298°
k
11,600
38.93
TK
298
dID
38.93ID
dVD
Flipping the result to define a resistance ratio (R V/I) gives us
dVD
0.026
dID
ID
or
26 mV
rd ID
(1.7)
Ge,Si
1.7
Resistance Levels
21
p n
The significance of Eq. (1.7) must be clearly understood. It implies that the dynamic
resistance can be found simply by substituting the quiescent value of the diode current into the equation. There is no need to have the characteristics available or to
worry about sketching tangent lines as defined by Eq. (1.6). It is important to keep
in mind, however, that Eq. (1.7) is accurate only for values of ID in the vertical-rise
section of the curve. For lesser values of ID, 2 (silicon) and the value of rd obtained must be multiplied by a factor of 2. For small values of ID below the knee of
the curve, Eq. (1.7) becomes inappropriate.
All the resistance levels determined thus far have been defined by the p-n junction and do not include the resistance of the semiconductor material itself (called body
resistance) and the resistance introduced by the connection between the semiconductor material and the external metallic conductor (called contact resistance). These additional resistance levels can be included in Eq. (1.7) by adding resistance denoted
by rB as appearing in Eq. (1.8). The resistance r d, therefore, includes the dynamic resistance defined by Eq. 1.7 and the resistance rB just introduced.
26 mV
r d ID
rB
ohms
(1.8)
The factor rB can range from typically 0.1 for high-power devices to 2 for
some low-power, general-purpose diodes. For Example 1.2 the ac resistance at 25 mA
was calculated to be 2 . Using Eq. (1.7), we have
26 mV
26 mV
rd 1.04 ID
25 mA
The difference of about 1 could be treated as the contribution of rB.
For Example 1.2 the ac resistance at 2 mA was calculated to be 27.5 . Using
Eq. (1.7) but multiplying by a factor of 2 for this region (in the knee of the curve
2),
26 mV
26 mV
rd 2 2 2(13 ) 26 ID
2 mA
The difference of 1.5 could be treated as the contribution due to rB.
In reality, determining rd to a high degree of accuracy from a characteristic curve
using Eq. (1.6) is a difficult process at best and the results have to be treated with a
grain of salt. At low levels of diode current the factor rB is normally small enough
compared to rd to permit ignoring its impact on the ac diode resistance. At high levels of current the level of rB may approach that of rd, but since there will frequently
be other resistive elements of a much larger magnitude in series with the diode we
will assume in this book that the ac resistance is determined solely by rd and the impact of rB will be ignored unless otherwise noted. Technological improvements of recent years suggest that the level of rB will continue to decrease in magnitude and
eventually become a factor that can certainly be ignored in comparison to rd.
The discussion above has centered solely on the forward-bias region. In the reverse-bias region we will assume that the change in current along the Is line is nil
from 0 V to the Zener region and the resulting ac resistance using Eq. (1.6) is sufficiently high to permit the open-circuit approximation.
Average AC Resistance
If the input signal is sufficiently large to produce a broad swing such as indicated in
Fig. 1.30, the resistance associated with the device for this region is called the average ac resistance. The average ac resistance is, by definition, the resistance deter22
Chapter 1
Semiconductor Diodes
p n
I D (mA)
20
15
∆ Id
10
5
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1
VD (V)
∆Vd
Figure 1.30 Determining the average ac resistance between indicated limits.
mined by a straight line drawn between the two intersections established by the maximum and minimum values of input voltage. In equation form (note Fig. 1.30),
Vd
rav Id
(1.9)
pt. to pt.
For the situation indicated by Fig. 1.30,
Id 17 mA 2 mA 15 mA
and
Vd 0.725 V 0.65 V 0.075 V
with
Vd
0.075 V
rav 5 Id
15 mA
If the ac resistance (rd) were determined at ID 2 mA its value would be more
than 5 , and if determined at 17 mA it would be less. In between the ac resistance
would make the transition from the high value at 2 mA to the lower value at 17 mA.
Equation (1.9) has defined a value that is considered the average of the ac values from
2 to 17 mA. The fact that one resistance level can be used for such a wide range of
the characteristics will prove quite useful in the definition of equivalent circuits for a
diode in a later section.
As with the dc and ac resistance levels, the lower the level of currents used to
determine the average resistance the higher the resistance level.
Summary Table
Table 1.2 was developed to reinforce the important conclusions of the last few pages
and to emphasize the differences among the various resistance levels. As indicated
earlier, the content of this section is the foundation for a number of resistance calculations to be performed in later sections and chapters.
1.7
Resistance Levels
23
p n
TABLE 1.2 Resistance Levels
Type
Equation
DC or static
VD
RD ID
AC or
dynamic
Average ac
V
26 mV
rd d Id
ID
Vd
rav Id pt. to pt.
Special
Characteristics
Graphical
Determination
Defined as a
point on the
characteristics
Defined by a
tangent line at
the Q-point
Defined by a straight
line between limits
of operation
1.8 DIODE EQUIVALENT CIRCUITS
An equivalent circuit is a combination of elements properly chosen to best
represent the actual terminal characteristics of a device, system, or such in a
particular operating region.
In other words, once the equivalent circuit is defined, the device symbol can be
removed from a schematic and the equivalent circuit inserted in its place without severely affecting the actual behavior of the system. The result is often a network that
can be solved using traditional circuit analysis techniques.
Piecewise-Linear Equivalent Circuit
One technique for obtaining an equivalent circuit for a diode is to approximate the
characteristics of the device by straight-line segments, as shown in Fig. 1.31. The resulting equivalent circuit is naturally called the piecewise-linear equivalent circuit. It
should be obvious from Fig. 1.31 that the straight-line segments do not result in an exact duplication of the actual characteristics, especially in the knee region. However,
the resulting segments are sufficiently close to the actual curve to establish an equivalent circuit that will provide an excellent first approximation to the actual behavior of
the device. For the sloping section of the equivalence the average ac resistance as introduced in Section 1.7 is the resistance level appearing in the equivalent circuit of Fig.
1.32 next to the actual device. In essence, it defines the resistance level of the device
when it is in the “on” state. The ideal diode is included to establish that there is only
one direction of conduction through the device, and a reverse-bias condition will re24
Chapter 1
Semiconductor Diodes
p n
Figure 1.31 Defining the
piecewise-linear equivalent circuit
using straight-line segments
to approximate the
characteristic curve.
+
VD
+
VD
VT
–
ID
ID
0.7 V
r av
–
Ideal diode
10 Ω
Figure 1.32 Components of the piecewise-linear equivalent circuit.
sult in the open-circuit state for the device. Since a silicon semiconductor diode does
not reach the conduction state until VD reaches 0.7 V with a forward bias (as shown
in Fig. 1.31), a battery VT opposing the conduction direction must appear in the equivalent circuit as shown in Fig. 1.32. The battery simply specifies that the voltage across
the device must be greater than the threshold battery voltage before conduction through
the device in the direction dictated by the ideal diode can be established. When conduction is established the resistance of the diode will be the specified value of rav.
Keep in mind, however, that VT in the equivalent circuit is not an independent
voltage source. If a voltmeter is placed across an isolated diode on the top of a lab
bench, a reading of 0.7 V will not be obtained. The battery simply represents the horizontal offset of the characteristics that must be exceeded to establish conduction.
The approximate level of rav can usually be determined from a specified operating point on the specification sheet (to be discussed in Section 1.9). For instance, for
a silicon semiconductor diode, if IF 10 mA (a forward conduction current for the
diode) at VD 0.8 V, we know for silicon that a shift of 0.7 V is required before the
characteristics rise and
Vd
rav Id
0.8 V 0.7 V
0.1 V
10 1
0
m
A
0
m
A
1
0 mA
pt. to pt.
as obtained for Fig. 1.30.
Simplified Equivalent Circuit
For most applications, the resistance rav is sufficiently small to be ignored in comparison to the other elements of the network. The removal of rav from the equivalent
1.8
Diode Equivalent Circut
25
p n
ID
+
VT = 0.7 V
r av = 0 Ω
ID
0
–
VD
Ideal diode
V T = 0.7 V V D
Figure 1.33 Simplified equivalent circuit for the silicon semiconductor diode.
circuit is the same as implying that the characteristics of the diode appear as shown
in Fig. 1.33. Indeed, this approximation is frequently employed in semiconductor circuit analysis as demonstrated in Chapter 2. The reduced equivalent circuit appears in
the same figure. It states that a forward-biased silicon diode in an electronic system
under dc conditions has a drop of 0.7 V across it in the conduction state at any level
of diode current (within rated values, of course).
Ideal Equivalent Circuit
Now that rav has been removed from the equivalent circuit let us take it a step further
and establish that a 0.7-V level can often be ignored in comparison to the applied
voltage level. In this case the equivalent circuit will be reduced to that of an ideal
diode as shown in Fig. 1.34 with its characteristics. In Chapter 2 we will see that this
approximation is often made without a serious loss in accuracy.
In industry a popular substitution for the phrase “diode equivalent circuit” is diode
model—a model by definition being a representation of an existing device, object,
system, and so on. In fact, this substitute terminology will be used almost exclusively
in the chapters to follow.
Figure 1.34 Ideal diode and its characteristics.
Summary Table
For clarity, the diode models employed for the range of circuit parameters and applications are provided in Table 1.3 with their piecewise-linear characteristics. Each will
be investigated in greater detail in Chapter 2. There are always exceptions to the general rule, but it is fairly safe to say that the simplified equivalent model will be employed most frequently in the analysis of electronic systems while the ideal diode is
frequently applied in the analysis of power supply systems where larger voltages are
encountered.
26
Chapter 1
Semiconductor Diodes
p n
TABLE 1.3 Diode Equivalent Circuits (Models)
Type
Conditions
Model
Characteristics
Piecewise-linear model
Simplified model
Rnetwork rav
Ideal device
Rnetwork rav
Enetwork VT
1.9 DIODE SPECIFICATION SHEETS
Data on specific semiconductor devices are normally provided by the manufacturer
in one of two forms. Most frequently, it is a very brief description limited to perhaps
one page. Otherwise, it is a thorough examination of the characteristics using graphs,
artwork, tables, and so on. In either case, there are specific pieces of data that must
be included for proper utilization of the device. They include:
1. The forward voltage VF (at a specified current and temperature)
2. The maximum forward current IF (at a specified temperature)
3. The reverse saturation current IR (at a specified voltage and temperature)
4. The reverse-voltage rating [PIV or PRV or V(BR), where BR comes from the term
“breakdown” (at a specified temperature)]
5. The maximum power dissipation level at a particular temperature
6. Capacitance levels (as defined in Section 1.10)
7. Reverse recovery time trr (as defined in Section 1.11)
8. Operating temperature range
Depending on the type of diode being considered, additional data may also be
provided, such as frequency range, noise level, switching time, thermal resistance levels, and peak repetitive values. For the application in mind, the significance of the
data will usually be self-apparent. If the maximum power or dissipation rating is also
provided, it is understood to be equal to the following product:
PDmax VD ID
(1.10)
where ID and VD are the diode current and voltage at a particular point of operation.
1.9
Diode Specification Sheets
27
p n
If we apply the simplified model for a particular application (a common occurrence), we can substitute VD VT 0.7 V for a silicon diode in Eq. (1.10) and determine the resulting power dissipation for comparison against the maximum power
rating. That is,
Pdissipated (0.7 V)ID
Figure 1.35 Electrical characteristics of a high-voltage, low-leakage diode.
28
Chapter 1
Semiconductor Diodes
(1.11)
p n
An exact copy of the data provided for a high-voltage/low-leakage diode appears
in Figs. 1.35 and 1.36. This example would represent the expanded list of data and
characteristics. The term rectifier is applied to a diode when it is frequently used in
a rectification process to be described in Chapter 2.
Figure 1.36 Terminal characteristics of a high-voltage diode.
1.9
Diode Specification Sheets
29
p n
Specific areas of the specification sheet have been highlighted in blue with a letter identification corresponding with the following description:
A: The minimum reverse-bias voltage (PIVs) for a diode at a specified reverse
saturation current.
B: Temperature characteristics as indicated. Note the use of the Celsius scale and
the wide range of utilization [recall that 32°F 0°C freezing (H2O) and
212°F 100°C boiling (H2O)].
C: Maximum power dissipation level PD VDID 500 mW. The maximum
power rating decreases at a rate of 3.33 mW per degree increase in temperature above room temperature (25°C), as clearly indicated by the power
derating curve of Fig. 1.36.
D: Maximum continuous forward current IFmax 500 mA (note IF versus temperature in Fig. 1.36).
E: Range of values of VF at IF 200 mA. Note that it exceeds VT 0.7 V for
both devices.
F: Range of values of VF at IF 1.0 mA. Note in this case how the upper limits surround 0.7 V.
G: At VR 20 V and a typical operating temperature IR 500 nA 0.5 A,
while at a higher reverse voltage IR drops to 5 nA 0.005 A.
H: The capacitance level between terminals is about 8 pF for the diode at VR VD 0 V (no-bias) and an applied frequency of 1 MHz.
I: The reverse recovery time is 3 s for the list of operating conditions.
A number of the curves of Fig. 1.36 employ a log scale. A brief investigation of
Section 11.2 should help with the reading of the graphs. Note in the top left figure
how VF increased from about 0.5 V to over 1 V as IF increased from 10 A to over
100 mA. In the figure below we find that the reverse saturation current does change
slightly with increasing levels of VR but remains at less than 1 nA at room temperature up to VR 125 V. As noted in the adjoining figure, however, note how quickly
the reverse saturation current increases with increase in temperature (as forecasted
earlier).
In the top right figure note how the capacitance decreases with increase in reversebias voltage, and in the figure below note that the ac resistance (rd) is only about
1 at 100 mA and increases to 100 at currents less than 1 mA (as expected from
the discussion of earlier sections).
The average rectified current, peak repetitive forward current, and peak forward
surge current as they appear on the specification sheet are defined as follows:
1. Average rectified current. A half-wave-rectified signal (described in Section 2.8)
has an average value defined by Iav 0.318Ipeak. The average current rating is
lower than the continuous or peak repetitive forward currents because a half-wave
current waveform will have instantaneous values much higher than the average
value.
2. Peak repetitive forward current. This is the maximum instantaneous value of repetitive forward current. Note that since it is at this level for a brief period of time,
its level can be higher than the continuous level.
3. Peak forward surge current. On occasion during turn-on, malfunctions, and so on,
there will be very high currents through the device for very brief intervals of time
(that are not repetitive). This rating defines the maximum value and the time interval for such surges in current level.
30
Chapter 1
Semiconductor Diodes
p n
The more one is exposed to specification sheets, the “friendlier” they will become,
especially when the impact of each parameter is clearly understood for the application under investigation.
1.10 TRANSITION AND DIFFUSION
CAPACITANCE
Electronic devices are inherently sensitive to very high frequencies. Most shunt capacitive effects that can be ignored at lower frequencies because the reactance XC 1/2 f C is very large (open-circuit equivalent). This, however, cannot be ignored at
very high frequencies. XC will become sufficiently small due to the high value of f to
introduce a low-reactance “shorting” path. In the p-n semiconductor diode, there are
two capacitive effects to be considered. Both types of capacitance are present in the
forward- and reverse-bias regions, but one so outweighs the other in each region that
we consider the effects of only one in each region.
In the reverse-bias region we have the transition- or depletion-region capacitance (CT), while in the forward-bias region we have the diffusion (CD ) or
storage capacitance.
Recall that the basic equation for the capacitance of a parallel-plate capacitor is
defined by C A/d, where is the permittivity of the dielectric (insulator) between
the plates of area A separated by a distance d. In the reverse-bias region there is a depletion region (free of carriers) that behaves essentially like an insulator between the
layers of opposite charge. Since the depletion width (d) will increase with increased
reverse-bias potential, the resulting transition capacitance will decrease, as shown in
Fig. 1.37. The fact that the capacitance is dependent on the applied reverse-bias potential has application in a number of electronic systems. In fact, in Chapter 20 a
diode will be introduced whose operation is wholly dependent on this phenomenon.
Although the effect described above will also be present in the forward-bias region, it is overshadowed by a capacitance effect directly dependent on the rate at
which charge is injected into the regions just outside the depletion region. The result
is that increased levels of current will result in increased levels of diffusion capacitance. However, increased levels of current result in reduced levels of associated resistance (to be demonstrated shortly), and the resulting time constant ( RC ), which
is very important in high-speed applications, does not become excessive.
C (pF)
15
10
C T)
Reverse-bias (C
5
CD )
Forward-bias (C
(V)
–25
–20
–15
–10
–5
0
0.25
0.5
Figure 1.37 Transition and diffusion capacitance versus applied bias for a
silicon diode.
1.10 Transition and Diffusion Capacitance
31
p n
The capacitive effects described above are represented by a capacitor in parallel
with the ideal diode, as shown in Fig. 1.38. For low- or mid-frequency applications
(except in the power area), however, the capacitor is normally not included in the
diode symbol.
Figure 1.38 Including the effect
of the transition or diffusion
capacitance on the semiconductor
diode.
1.11 REVERSE RECOVERY TIME
There are certain pieces of data that are normally provided on diode specification
sheets provided by manufacturers. One such quantity that has not been considered yet
is the reverse recovery time, denoted by trr . In the forward-bias state it was shown
earlier that there are a large number of electrons from the n-type material progressing through the p-type material and a large number of holes in the n-type—a requirement for conduction. The electrons in the p-type and holes progressing through
the n-type material establish a large number of minority carriers in each material. If
the applied voltage should be reversed to establish a reverse-bias situation, we would
ideally like to see the diode change instantaneously from the conduction state to the
nonconduction state. However, because of the large number of minority carriers in
each material, the diode current will simply reverse as shown in Fig. 1.39 and stay at
this measurable level for the period of time ts (storage time) required for the minority carriers to return to their majority-carrier state in the opposite material. In essence,
the diode will remain in the short-circuit state with a current Ireverse determined by
the network parameters. Eventually, when this storage phase has passed, the current
will reduce in level to that associated with the nonconduction state. This second period of time is denoted by tt (transition interval). The reverse recovery time is the sum
of these two intervals: trr ts tt. Naturally, it is an important consideration in highspeed switching applications. Most commercially available switching diodes have a
trr in the range of a few nanoseconds to 1 s. Units are available, however, with a trr
of only a few hundred picoseconds (1012).
ID
Change of state (on
required at t = t 1
I forward
off)
Desired response
t1
t
I reverse
ts
tt
t rr
Figure 1.39 Defining the
reverse recovery time.
1.12 SEMICONDUCTOR DIODE
NOTATION
The notation most frequently used for semiconductor diodes is provided in Fig. 1.40.
For most diodes any marking such as a dot or band, as shown in Fig. 1.40, appears
at the cathode end. The terminology anode and cathode is a carryover from vacuumtube notation. The anode refers to the higher or positive potential, and the cathode
refers to the lower or negative terminal. This combination of bias levels will result in
a forward-bias or “on” condition for the diode. A number of commercially available
semiconductor diodes appear in Fig. 1.41. Some details of the actual construction of
devices such as those appearing in Fig. 1.41 are provided in Chapters 12 and 20.
32
Chapter 1
Semiconductor Diodes
p n
Anode
p
n
or •, K, etc.
Cathode
Figure 1.40 Semiconductor diode notation.
Figure 1.41 Various types of junction diodes. [(a) Courtesy of Motorola Inc.;
and (b) and (c) Courtesy International Rectifier Corporation.]
1.13 DIODE TESTING
The condition of a semiconductor diode can be determined quickly using (1) a digital display meter (DDM) with a diode checking function, (2) the ohmmeter section of
a multimeter, or (3) a curve tracer.
Diode Checking Function
A digital display meter with a diode checking capability appears in Fig. 1.42. Note
the small diode symbol as the bottom option of the rotating dial. When set in this position and hooked up as shown in Fig. 1.43a, the diode should be in the “on” state
and the display will provide an indication of the forward-bias voltage such as 0.67 V
(for Si). The meter has an internal constant current source (about 2 mA) that will define the voltage level as indicated in Fig. 1.43b. An OL indication with the hookup
of Fig. 1.43a reveals an open (defective) diode. If the leads are reversed, an OL indication should result due to the expected open-circuit equivalence for the diode. In
general, therefore, an OL indication in both directions is an indication of an open or
defective diode.
1.13
Diode Testing
33
p n
Figure 1.42 Digital display
meter with diode checking
capability. (Courtesy
Computronics Technology, Inc.)
Figure 1.43 Checking a diode
in the forward-bias state.
Ohmmeter Testing
In Section 1.7 we found that the forward-bias resistance of a semiconductor diode is
quite low compared to the reverse-bias level. Therefore, if we measure the resistance
of a diode using the connections indicated in Fig. 1.44a, we can expect a relatively
low level. The resulting ohmmeter indication will be a function of the current established through the diode by the internal battery (often 1.5 V) of the ohmmeter circuit.
The higher the current, the less the resistance level. For the reverse-bias situation the
reading should be quite high, requiring a high resistance scale on the meter, as indicated in Fig. 1.44b. A high resistance reading in both directions obviously indicates
an open (defective device) condition, while a very low resistance reading in both directions will probably indicate a shorted device.
Curve Tracer
Figure 1.44 Checking a diode
with an ohmmeter.
The curve tracer of Fig. 1.45 can display the characteristics of a host of devices, including the semiconductor diode. By properly connecting the diode to the test panel
at the bottom center of the unit and adjusting the controls, the display of Fig. 1.46
34
Chapter 1
Semiconductor Diodes
p n
Figure 1.45 Curve tracer.
(Courtesy of Tektronix, Inc.)
Figure 1.46 Curve tracer
response to 1N4007 silicon diode.
can be obtained. Note that the vertical scaling is 1 mA/div, resulting in the levels indicated. For the horizontal axis the scaling is 100 mV/div, resulting in the voltage levels indicated. For a 2-mA level as defined for a DDM, the resulting voltage would be
about 625 mV 0.625 V. Although the instrument initially appears quite complex,
the instruction manual and a few moments of exposure will reveal that the desired results can usually be obtained without an excessive amount of effort and time. The
same instrument will appear on more than one occasion in the chapters to follow as
we investigate the characteristics of the variety of devices.
1.14 ZENER DIODES
The Zener region of Fig. 1.47 was discussed in some detail in Section 1.6. The characteristic drops in an almost vertical manner at a reverse-bias potential denoted VZ.
The fact that the curve drops down and away from the horizontal axis rather than up
and away for the positive VD region reveals that the current in the Zener region has
a direction opposite to that of a forward-biased diode.
1.14
Zener Diodes
Figure 1.47 Reviewing the
Zener region.
35
p n
Figure 1.48 Conduction direction: (a) Zener diode; (b) semiconductor diode.
Figure 1.49 Zener equivalent
circuit: (a) complete; (b) approximate.
This region of unique characteristics is employed in the design of Zener diodes,
which have the graphic symbol appearing in Fig. 1.48a. Both the semiconductor diode
and zener diode are presented side by side in Fig. 1.48 to ensure that the direction of
conduction of each is clearly understood together with the required polarity of the applied voltage. For the semiconductor diode the “on” state will support a current in the
direction of the arrow in the symbol. For the Zener diode the direction of conduction
is opposite to that of the arrow in the symbol as pointed out in the introduction to this
section. Note also that the polarity of VD and VZ are the same as would be obtained
if each were a resistive element.
The location of the Zener region can be controlled by varying the doping levels.
An increase in doping, producing an increase in the number of added impurities, will
decrease the Zener potential. Zener diodes are available having Zener potentials of
1.8 to 200 V with power ratings from 14 to 50 W. Because of its higher temperature
and current capability, silicon is usually preferred in the manufacture of Zener diodes.
The complete equivalent circuit of the Zener diode in the Zener region includes
a small dynamic resistance and dc battery equal to the Zener potential, as shown in
Fig. 1.49. For all applications to follow, however, we shall assume as a first approximation that the external resistors are much larger in magnitude than the Zener-equivalent resistor and that the equivalent circuit is simply the one indicated in Fig. 1.49b.
A larger drawing of the Zener region is provided in Fig. 1.50 to permit a description of the Zener nameplate data appearing in Table 1.4 for a 10-V, 500-mW, 20%
diode. The term nominal associated with VZ indicates that it is a typical average value.
Since this is a 20% diode, the Zener potential can be expected to vary as 10 V 20%
Figure 1.50 Zener test
characteristics.
TABLE 1.4 Electrical Characteristics (25°C Ambient Temperature Unless Otherwise Noted)
Zener
Voltage
Nominal,
VZ
(V)
Test
Current,
IZT
(mA)
Max
Dynamic
Impedance,
ZZT at IZT
()
10
12.5
8.5
36
Chapter 1
Maximum
Knee
Impedance,
ZZK at IZK
() (mA)
700
Maximum
Reverse
Current,
IR at VR
( A)
Test
Voltage,
VR
(V)
Maximum
Regulator
Current,
IZM
(mA)
10
7.2
32
0.25
Semiconductor Diodes
Typical
Temperature
Coefficient
(%/°C)
0.072
p n
or from 8 to 12 V in its range of application. Also available are 10% and 5% diodes
with the same specifications. The test current IZT is the current defined by the 14 power
level, and ZZT is the dynamic impedance at this current level. The maximum knee impedance occurs at the knee current of IZK. The reverse saturation current is provided
at a particular potential level, and IZM is the maximum current for the 20% unit.
The temperature coefficient reflects the percent change in VZ with temperature. It
is defined by the equation
VZ
100%
TC VZ (T1 T0)
%/°C
(1.12)
where VZ is the resulting change in Zener potential due to the temperature variation. Note in Fig. 1.51a that the temperature coefficient can be positive, negative, or
even zero for different Zener levels. A positive value would reflect an increase in VZ
with an increase in temperature, while a negative value would result in a decrease in
value with increase in temperature. The 24-V, 6.8-V, and 3.6-V levels refer to three
Zener diodes having these nominal values within the same family of Zeners. The curve
for the 10-V Zener would naturally lie between the curves of the 6.8-V and 24-V devices. Returning to Eq. (1.12), T0 is the temperature at which VZ is provided (normally room temperature—25°C), and T1 is the new level. Example 1.3 will demonstrate the use of Eq. (1.12).
Temperature coefficient
versus Zener current
Dynamic impedence
versus Zener current
+0.08
24 V
+0.04
6.8 V
Dynamic impedance, ZZ – ( Ω )
Temperature coefficient – TC (%/˚C)
+0.12
0
– 0.04
3.6 V
– 0.08
1 kΩ
500
200
100
50
3.6 V
20
10
5
24 V
2
– 0.12
0.01 0.05 0.1
0.5 1
5 10
50 100
1
0.1 0.2 0.5 1
Zener current, IZ – (mA)
2
6.8 V
5 10 20 50 100
Zener current, IZ – (mA)
(a)
(b)
Figure 1.51 Electrical characteristics for a 10-V, 500-mW Zener diode.
Determine the nominal voltage for the Zener diode of Table 1.4 at a temperature of
100°C.
EXAMPLE 1.3
Solution
From Eq. 1.12,
T VZ
VZ C
(T1 T0)
100
1.14
Zener Diodes
37
p n
Substitution values from Table 1.4 yield
(0.072)(10 V)
VZ (100°C 25°C)
100
(0.0072)(75)
0.54 V
and because of the positive temperature coefficient, the new Zener potential, defined
by VZ, is
VZ VZ
0.54 V
10.54 V
The variation in dynamic impedance (fundamentally, its series resistance) with
current appears in Fig. 1.51b. Again, the 10-V Zener appears between the 6.8-V and
24-V Zeners. Note that the heavier the current (or the farther up the vertical rise you
are in Fig. 1.47), the less the resistance value. Also note that as you drop below the
knee of the curve, the resistance increases to significant levels.
The terminal identification and the casing for a variety of Zener diodes appear in
Fig. 1.52. Figure 1.53 is an actual photograph of a variety of Zener devices. Note that
their appearance is very similar to the semiconductor diode. A few areas of application for the Zener diode will be examined in Chapter 2.
Figure 1.52 Zener terminal
identification and symbols.
Figure 1.53 Zener diodes.
(Courtesy Siemens Corporation.)
1.15 LIGHT-EMITTING DIODES
The increasing use of digital displays in calculators, watches, and all forms of instrumentation has contributed to the current extensive interest in structures that will
emit light when properly biased. The two types in common use today to perform this
function are the light-emitting diode (LED) and the liquid-crystal display (LCD). Since
the LED falls within the family of p-n junction devices and will appear in some of
38
Chapter 1
Semiconductor Diodes
p n
the networks in the next few chapters, it will be introduced in this chapter. The LCD
display is described in Chapter 20.
As the name implies, the light-emitting diode (LED) is a diode that will give off
visible light when it is energized. In any forward-biased p-n junction there is, within
the structure and primarily close to the junction, a recombination of holes and electrons. This recombination requires that the energy possessed by the unbound free electron be transferred to another state. In all semiconductor p-n junctions some of this
energy will be given off as heat and some in the form of photons. In silicon and germanium the greater percentage is given up in the form of heat and the emitted light
is insignificant. In other materials, such as gallium arsenide phosphide (GaAsP) or
gallium phosphide (GaP), the number of photons of light energy emitted is sufficient
to create a very visible light source.
The process of giving off light by applying an electrical source of energy is
called electroluminescence.
As shown in Fig. 1.54 with its graphic symbol, the conducting surface connected
to the p-material is much smaller, to permit the emergence of the maximum number
of photons of light energy. Note in the figure that the recombination of the injected
carriers due to the forward-biased junction results in emitted light at the site of recombination. There may, of course, be some absorption of the packages of photon energy in the structure itself, but a very large percentage are able to leave, as shown in
the figure.
Figure 1.54 (a) Process of
electroluminescence in the LED;
(b) graphic symbol.
The appearance and characteristics of a subminiature high-efficiency solid-state
lamp manufactured by Hewlett-Packard appears in Fig. 1.55. Note in Fig. 1.55b that
the peak forward current is 60 mA, with 20 mA the typical average forward current.
The test conditions listed in Fig. 1.55c, however, are for a forward current of 10 mA.
The level of VD under forward-bias conditions is listed as VF and extends from 2.2
to 3 V. In other words, one can expect a typical operating current of about 10 mA at
2.5 V for good light emission.
Two quantities yet undefined appear under the heading Electrical/Optical Characteristics at TA 25°C. They are the axial luminous intensity (IV) and the luminous
efficacy ( v). Light intensity is measured in candela. One candela emits a light flux
of 4 lumens and establishes an illumination of 1 footcandle on a 1-ft2 area 1 ft from
the light source. Even though this description may not provide a clear understanding
of the candela as a unit of measure, its level can certainly be compared between similar devices. The term efficacy is, by definition, a measure of the ability of a device
to produce a desired effect. For the LED this is the ratio of the number of lumens
generated per applied watt of electrical energy. The relative efficiency is defined by
1.15
Light-Emitting Diodes
39
p n
the luminous intensity per unit current, as shown in Fig. 1.55g. The relative intensity
of each color versus wavelength appears in Fig. 1.55d.
Since the LED is a p-n junction device, it will have a forward-biased characteristic
(Fig. 1.55e) similar to the diode response curves. Note the almost linear increase in relative luminous intensity with forward current (Fig. 1.55f). Figure 1.55h reveals that the
longer the pulse duration at a particular frequency, the lower the permitted peak current
(after you pass the break value of tp). Figure 1.55i simply reveals that the intensity is
greater at 0° (or head on) and the least at 90° (when you view the device from the side).
Figure 1.55 Hewlett-Packard subminiature high-efficiency red solid-state lamp: (a) appearance;
(b) absolute maximum ratings; (c) electrical/optical characteristics; (d) relative intensity versus wavelength; (e) forward current versus forward voltage; (f) relative luminous intensity versus forward current; (g) relative efficiency versus peak current; (h) maximum peak current versus pulse duration;
(i) relative luminous intensity versus angular displacement. (Courtesy Hewlett-Packard Corporation.)
40
Chapter 1
Semiconductor Diodes
p n
1.0
Relative intensity
Green
TA = 25˚C
GaAsP Red
Yellow
High efficiency
Red
0.5
0
500
550
600
650
Wavelength–nm
700
750
(d)
20
15
10
5
2.0
1.0
0
0
0.5
1.0 1.5 2.0 2.5
3.0
0
5
VF – Forward voltage – V
20
0
10
20
30
40
50
Ipeak – Peak current – mA
(f)
(g)
60
5
tp
4
20˚
40˚
0.6
Hz
100
Hz
300
z
1 kH
10
100
1000
tp – Pulse duration – µs
0.8
60˚
70˚
80˚
1
1.0
0˚
50˚
T
z
3 kH
z
10 kH
–
2
10˚
30˚
3
z
30 kH
Hz
Idc max
15
6
100 k
Ratio of maximum tolerable
peak current
to maximum tolerable
dc current
10
1.6
1.5
1.4
1.3
1.2
1.1
1.0
0.9
0.8
0.7
0.6
IF – Forward current – mA
(e)
Ipeak max
Relative efficiency
(normalized at 10 mA dc)
Relative luminous intensity
(normalized at 10 mA)
IF – Forward current – mA
TA = 25˚C
0
TA = 25˚C
3.0
0.4
0.2
90˚
10,000
(h)
20˚ 40˚
60˚ 80˚ 100˚
(i)
Figure 1.55 Continued.
1.15
Light-Emitting Diodes
41
p n
LED displays are available today in many different sizes and shapes. The lightemitting region is available in lengths from 0.1 to 1 in. Numbers can be created by
segments such as shown in Fig. 1.56. By applying a forward bias to the proper p-type
material segment, any number from 0 to 9 can be displayed.
Figure 1.56 Litronix segment display.
There are also two-lead LED lamps that contain two LEDs, so that a reversal in
biasing will change the color from green to red, or vice versa. LEDs are presently
available in red, green, yellow, orange, and white, and white with blue soon to be
commercially available. In general, LEDs operate at voltage levels from 1.7 to 3.3 V,
which makes them completely compatible with solid-state circuits. They have a fast
response time (nanoseconds) and offer good contrast ratios for visibility. The power
requirement is typically from 10 to 150 mW with a lifetime of 100,000 hours. Their
semiconductor construction adds a significant ruggedness factor.
1.16 DIODE ARRAYS—INTEGRATED
CIRCUITS
The unique characteristics of integrated circuits will be introduced in Chapter 12.
However, we have reached a plateau in our introduction to electronic circuits that permits at least a surface examination of diode arrays in the integrated-circuit package.
You will find that the integrated circuit is not a unique device with characteristics totally different from those we examine in these introductory chapters. It is simply a
packaging technique that permits a significant reduction in the size of electronic systems. In other words, internal to the integrated circuit are systems and discrete devices that were available long before the integrated circuit as we know it today became a reality.
One possible array appears in Fig. 1.57. Note that eight diodes are internal to the
diode array. That is, in the container shown in Fig. 1.58 there are diodes set in a single silicon wafer that have all the anodes connected to pin 1 and the cathodes of each
to pins 2 through 9. Note in the same figure that pin 1 can be determined as being to
the left of the small projection in the case if we look from the bottom toward the case.
The other numbers then follow in sequence. If only one diode is to be used, then only
pins 1 and 2 (or any number from 3 to 9) would be used. The remaining diodes would
be left hanging and not affect the network to which pins 1 and 2 are connected.
Another diode array appears in Fig. 1.59 (see page 44). In this case the package
is different but the numbering sequence appears in the outline. Pin 1 is the pin directly above the small indentation as you look down on the device.
42
Chapter 1
Semiconductor Diodes
p n
Figure 1.57 Monolithic diode array.
1.17 PSPICE WINDOWS
The computer has now become such an integral part of the electronics industry that
the capabilities of this working “tool” must be introduced at the earliest possible opportunity. For those students with no prior computer experience there is a common
initial fear of this seemingly complicated powerful system. With this in mind the computer analysis of this book was designed to make the computer system more “friendly”
by revealing the relative ease with which it can be applied to perform some very help1.16 Diode Arrays — Integrated Circuits
43
p n
Figure 1.58 Package outline
TO-96 for a diode array. All
dimensions are in inches.
TO-116-2 Outline
0.785"
Connection Diagrams
FSA2500M
1
7
1
8
14
0.271"
2
3
4
5
6
7
8
9
0.310"
10
0.200"
max.
Seating
plane
Notes:
Alloy 42 pins, tin plated
Gold plated pins available
Hermetically sealed ceramic
package
Figure 1.59 Monolithic diode array. All dimensions are in inches.
ful and special tasks in a minimum amount of time with a high degree of accuracy.
The content was written with the assumption that the reader has no prior computer
experience or exposure to the terminology to be applied. There is also no suggestion
that the content of this book is sufficient to permit a complete understanding of the
“hows” and “whys” that will surface. The purpose here is solely to introduce some
of the terminology, discuss a few of its capabilities, reveal the possibilities available,
touch on some of its limitations, and demonstrate its versatility with a number of carefully chosen examples.
In general, the computer analysis of electronic systems can take one of two approaches: using a language such as BASIC, Fortran, Pascal, or C; or utilizing a software package such as PSpice, MicroCap II, Breadboard, or Circuit Master, to name
a few. A language, through its symbolic notation, forms a bridge between the user
and the computer that permits a dialogue between the two for establishing the operations to be performed.
In earlier editions of this text, the chosen language was BASIC, primarily because
it uses a number of familiar words and phrases from the English language that in
themselves reveal the operation to be performed. When a language is employed to analyze a system, a program is developed that sequentially defines the operations to be
performed—in much the same order in which we perform the same analysis in longhand. As with the longhand approach, one wrong step and the result obtained can be
completely meaningless. Programs typically develop with time and application as
more efficient paths toward a solution become obvious. Once established in its “best”
form it can be cataloged for future use. The important advantage of the language approach is that a program can be tailored to meet all the special needs of the user. It
permits innovative “moves” by the user that can result in printouts of data in an informative and interesting manner.
The alternative approach referred to above utilizes a software package to perform
the desired investigation. A software package is a program written and tested over a
44
Chapter 1
Semiconductor Diodes
p n
period of time designed to perform a particular type of analysis or synthesis in an efficient manner with a high level of accuracy.
The package itself cannot be altered by the user, and its application is limited to the
operations built into the system. A user must adjust his or her desire for output information to the range of possibilities offered by the package. In addition, the user must
input information exactly as requested by the package or the data may be misinterpreted.
The software package chosen for this book is PSpice.* PSpice currently is available in
two forms: DOS and Windows. Although DOS format was the first introduced, the Windows version is the most popular today. The Windows version employed in this text is
8.0, the latest available. A photograph of a complete Design Center package appears in
Fig. 1.60 with the 8.0 CD-ROM version. It is also available in 3.5 diskettes. A more
sophisticated version referred to simply as SPICE is finding widespread application in
industry.
Figure 1.60 PSpice Design
package. (Courtesy of the
OrCAD-MicroSim Corporation.)
In total, therefore, a software package is “packaged” to perform a specific series
of calculations and operations and to provide the results in a defined format. A language permits an expanded level of flexibility but also fails to benefit from the extensive testing and research normally devoted to the development of a “trusted” package. The user must define which approach best fits the needs of the moment. Obviously,
if a package exists for the desired analysis or synthesis, it should be considered before turning to the many hours required to develop a reliable, efficient program. In
addition, one may acquire the data needed for a particular analysis from a software
package and then turn to a language to define the format of the output. In many ways,
the two approaches go hand in hand. If one is to depend on computer analysis on a
continuing basis, knowledge of the use and limits of both languages and software
packages is a necessity. The choice of which language or software package to become
familiar with is primarily a function of the area of investigation. Fortunately, however, a fluent knowledge of one language or a particular software package will usually help the user become familiar with other languages and software packages. There
is a similarity in purpose and procedures that ease the transition from one approach
to another.
When using PSpice Windows, the network is first drawn on the screen followed
by an analysis dictated by the needs of the user. This text will be using Version 8.0,
though the differences between this and earlier Windows versions are so few and
relatively minor for this level of application that one should not be concerned if using an earlier edition. The first step, of course, is to install PSpice into the hard-disk
*PSpice is a registered trademark of the OrCAD-MicroSim Corporation.
1.17 PSpice Windows
45
p n
memory of your computer following the directions provided by MicroSim. Next,
the Schematics screen must be obtained using a control mechanism such as
Windows 95. Once established, the elements for the network must be obtained and
placed on the screen to build the network. In this text, the procedure for each element
will be described following the discussion of the characteristics and analysis of each
device.
Since we have just finished covering the diode in detail, the procedure for finding the diodes stored in the library will be introduced along with the method for placing them on the screen. The next chapter will introduce the procedure for analyzing
a complete network with diodes using PSpice. There are several ways to proceed, but
the most direct path is to click on the picture symbol with the binoculars on the top
right of the schematics screen. As you bring the marker close to the box using the
mouse, a message Get New Part will be displayed. Left click on the symbol and a
Part Browser Basic dialog box will appear. By choosing Libraries, a Library
Browser dialog box will appear and the EVAL.slb library should be chosen. When
selected, all available parts in this library will appear in the Part listing. Next, scroll
the Part list and choose the D1N4148 diode. The result is that the Part Name will
appear above and the Description will indicate it is a diode. Once set, click OK and
the Part Browser Basic dialog box will reappear with the full review of the chosen
element. To place the device on the screen and close the dialog box, simply click on
the Place & Close option. The result is that the diode will appear on the screen and
can be put in place with a left click of the mouse. Once located, two labels will appear—one indicating how any diodes have been placed (D1, D2, D3, and so on) and
the other the name of the chosen diode (D1N4148). The same diode can be placed in
other places on the same screen by simply moving the pointer and left clicking the
mouse. The process can be ended by a single right click of the mouse. Any of the
diodes can be removed by simply clicking on them to make them red and pressing
the Delete key. If preferred, the Edit choice of the menu bar at the top of the screen
also can be chosen, followed by using the Delete command.
Another path for obtaining an element is to choose Draw on the menu bar, followed by Get New Part. Once chosen, the Part Browser Basic dialog box will appear as before and the same procedure can be followed. Now that we know the
D1N4148 diode exists, it can be obtained directly once the Part Browser Basic dialog box appears. Simply type D1N4148 in the Part Name box, followed by Place
& Close, and the diode will appear on the screen.
If a diode has to be moved, simply left click on it once, until it turns red. Then,
click on it again and hold the clicker down on the mouse. At the same time, move the
diode to any location you prefer and, when set, lift up on the clicker. Remember that
anything in red can be operated on. To remove the red status, simply remove the
pointer from the element and click it once. The diode will turn green and blue, indicating that its location and associated information is set in memory. For all the above
and for the chapters to follow, if you happen to have a monochromatic (black-andwhite) screen, you will simply have to remember whether the device is in the active
state.
If the label or parameters of the diode are to be changed, simply click on the element once (to make it red) and choose Edit, followed by Model. An Edit Model
dialog box will appear with a choice of changing the model reference (D1N4148),
the text associated with each parameter, or the parameters that define the characteristics of the diode.
As mentioned above, additional comments regarding use of the diode will be made
in the chapters to follow. For the moment, we are at least aware of how to find and
place an element on the screen. If time permits, review the other elements available
within the various libraries to prepare yourself for the work to follow.
46
Chapter 1
Semiconductor Diodes
p n
§ 1.2 Ideal Diode
PROBLEMS
1. Describe in your own words the meaning of the word ideal as applied to a device or system.
2. Describe in your own words the characteristics of the ideal diode and how they determine the
on and off states of the device. That is, describe why the short-circuit and open-circuit equivalents are appropriate.
3. What is the one important difference between the characteristics of a simple switch and those
of an ideal diode?
§ 1.3 Semiconductor Materials
4. In your own words, define semiconductor, resistivity, bulk resistance, and ohmic contact resistance.
5. (a) Using Table 1.1, determine the resistance of a silicon sample having an area of 1 cm2 and
a length of 3 cm.
(b) Repeat part (a) if the length is 1 cm and the area 4 cm2.
(c) Repeat part (a) if the length is 8 cm and the area 0.5 cm2.
(d) Repeat part (a) for copper and compare the results.
6. Sketch the atomic structure of copper and discuss why it is a good conductor and how its structure is different from germanium and silicon.
7. In your own words, define an intrinsic material, a negative temperature coefficient, and covalent bonding.
8. Consult your reference library and list three materials that have a negative temperature coefficient and three that have a positive temperature coefficient.
§ 1.4 Energy Levels
9. How much energy in joules is required to move a charge of 6 C through a difference in potential of 3 V?
10. If 48 eV of energy is required to move a charge through a potential difference of 12 V, determine the charge involved.
11. Consult your reference library and determine the level of Eg for GaP and ZnS, two semiconductor materials of practical value. In addition, determine the written name for each material.
§ 1.5 Extrinsic Materials—n- and p-Type
12. Describe the difference between n-type and p-type semiconductor materials.
13. Describe the difference between donor and acceptor impurities.
14. Describe the difference between majority and minority carriers.
15. Sketch the atomic structure of silicon and insert an impurity of arsenic as demonstrated for silicon in Fig. 1.9.
16. Repeat Problem 15 but insert an impurity of indium.
17. Consult your reference library and find another explanation of hole versus electron flow. Using both descriptions, describe in your own words the process of hole conduction.
§ 1.6 Semiconductor Diode
18. Describe in your own words the conditions established by forward- and reverse-bias conditions
on a p-n junction diode and how the resulting current is affected.
19. Describe how you will remember the forward- and reverse-bias states of the p-n junction diode.
That is, how you will remember which potential (positive or negative) is applied to which terminal?
20. Using Eq. (1.4), determine the diode current at 20°C for a silicon diode with Is 50 nA and
an applied forward bias of 0.6 V.
Problems
47
p n
21. Repeat Problem 20 for T 100°C (boiling point of water). Assume that Is has increased to
5.0 A.
22. (a) Using Eq. (1.4), determine the diode current at 20°C for a silicon diode with Is 0.1 A
at a reverse-bias potential of 10 V.
(b) Is the result expected? Why?
23. (a) Plot the function y ex for x from 0 to 5.
(b) What is the value of y ex at x 0?
(c) Based on the results of part (b), why is the factor 1 important in Eq. (1.4)?
24. In the reverse-bias region the saturation current of a silicon diode is about 0.1 A (T 20°C).
Determine its approximate value if the temperature is increased 40°C.
25. Compare the characteristics of a silicon and a germanium diode and determine which you would
prefer to use for most practical applications. Give some details. Refer to a manufacturer’s listing and compare the characteristics of a germanium and a silicon diode of similar maximum
ratings.
26. Determine the forward voltage drop across the diode whose characteristics appear in Fig. 1.24
at temperatures of 75°C, 25°C, 100°C, and 200°C and a current of 10 mA. For each temperature, determine the level of saturation current. Compare the extremes of each and comment
on the ratio of the two.
§ 1.7 Resistance Levels
27. Determine the static or dc resistance of the commercially available diode of Fig. 1.19 at a forward current of 2 mA.
28. Repeat Problem 26 at a forward current of 15 mA and compare results.
29. Determine the static or dc resistance of the commercially available diode of Fig. 1.19 at a reverse voltage of 10 V. How does it compare to the value determined at a reverse voltage of
30 V?
30. (a) Determine the dynamic (ac) resistance of the diode of Fig. 1.29 at a forward current of 10
mA using Eq. (1.6).
(b) Determine the dynamic (ac) resistance of the diode of Fig. 1.29 at a forward current of 10
mA using Eq. (1.7).
(c) Compare solutions of parts (a) and (b).
31. Calculate the dc and ac resistance for the diode of Fig. 1.29 at a forward current of 10 mA and
compare their magnitudes.
32. Using Eq. (1.6), determine the ac resistance at a current of 1 mA and 15 mA for the diode of
Fig. 1.29. Compare the solutions and develop a general conclusion regarding the ac resistance
and increasing levels of diode current.
33. Using Eq. (1.7), determine the ac resistance at a current of 1 mA and 15 mA for the diode of
Fig. 1.19. Modify the equation as necessary for low levels of diode current. Compare to the solutions obtained in Problem 32.
34. Determine the average ac resistance for the diode of Fig. 1.19 for the region between 0.6 and
0.9 V.
35. Determine the ac resistance for the diode of Fig. 1.19 at 0.75 V and compare to the average ac
resistance obtained in Problem 34.
§ 1.8 Diode Equivalent Circuits
36. Find the piecewise-linear equivalent circuit for the diode of Fig. 1.19. Use a straight line segment that intersects the horizontal axis at 0.7 V and best approximates the curve for the region
greater than 0.7 V.
37. Repeat Problem 36 for the diode of Fig. 1.29.
48
Chapter 1
Semiconductor Diodes
p n
§ 1.9 Diode Specification Sheets
* 38. Plot IF versus VF using linear scales for the diode of Fig. 1.36. Note that the provided graph
employs a log scale for the vertical axis (log scales are covered in sections 11.2 and 11.3).
39. Comment on the change in capacitance level with increase in reverse-bias potential for the diode
of Fig. 1.36.
40. Does the reverse saturation current of the diode of Fig. 1.36 change significantly in magnitude
for reverse-bias potentials in the range 25 to 100 V?
* 41. For the diode of Fig. 1.36 determine the level of IR at room temperature (25°C) and the boiling point of water (100°C). Is the change significant? Does the level just about double for every
10°C increase in temperature?
42. For the diode of Fig. 1.36 determine the maximum ac (dynamic) resistance at a forward current of 0.1, 1.5, and 20 mA. Compare levels and comment on whether the results support conclusions derived in earlier sections of this chapter.
43. Using the characteristics of Fig. 1.36, determine the maximum power dissipation levels for the
diode at room temperature (25°C) and 100°C. Assuming that VF remains fixed at 0.7 V, how
has the maximum level of IF changed between the two temperature levels?
44. Using the characteristics of Fig. 1.36, determine the temperature at which the diode current will
be 50% of its value at room temperature (25°C).
§ 1.10 Transition and Diffusion Capacitance
* 45. (a) Referring to Fig. 1.37, determine the transition capacitance at reverse-bias potentials of 25
and 10 V. What is the ratio of the change in capacitance to the change in voltage?
(b) Repeat part (a) for reverse-bias potentials of 10 and 1 V. Determine the ratio of the
change in capacitance to the change in voltage.
(c) How do the ratios determined in parts (a) and (b) compare? What does it tell you about
which range may have more areas of practical application?
46. Referring to Fig. 1.37, determine the diffusion capacitance at 0 and 0.25 V.
47. Describe in your own words how diffusion and transition capacitances differ.
48. Determine the reactance offered by a diode described by the characteristics of Fig. 1.37 at a
forward potential of 0.2 V and a reverse potential of 20 V if the applied frequency is 6 MHz.
§ 1.11 Reverse Recovery Time
49. Sketch the waveform for i of the network of Fig. 1.61 if tt 2ts and the total reverse recovery
time is 9 ns.
Figure 1.61 Problem 49
§ 1.14 Zener Diodes
50. The following characteristics are specified for a particular Zener diode: VZ 29 V, VR 16.8 V,
IZT 10 mA, IR 20 A, and IZM 40 mA. Sketch the characteristic curve in the manner
displayed in Fig. 1.50.
* 51. At what temperature will the 10-V Zener diode of Fig. 1.50 have a nominal voltage of 10.75 V?
(Hint: Note the data in Table 1.4.)
Problems
49
p n
52. Determine the temperature coefficient of a 5-V Zener diode (rated 25°C value) if the nominal
voltage drops to 4.8 V at a temperature of 100°C.
53. Using the curves of Fig. 1.51a, what level of temperature coefficient would you expect for a
20-V diode? Repeat for a 5-V diode. Assume a linear scale between nominal voltage levels and
a current level of 0.1 mA.
54. Determine the dynamic impedance for the 24-V diode at IZ 10 mA for Fig. 1.51b. Note that
it is a log scale.
* 55. Compare the levels of dynamic impedance for the 24-V diode of Fig. 1.51b at current levels of
0.2, 1, and 10 mA. How do the results relate to the shape of the characteristics in this region?
§ 1.15 Light-Emitting Diodes
56. Referring to Fig. 1.55e, what would appear to be an appropriate value of VT for this device?
How does it compare to the value of VT for silicon and germanium?
57. Using the information provided in Fig. 1.55, determine the forward voltage across the diode if
the relative luminous intensity is 1.5.
* 58. (a) What is the percent increase in relative efficiency of the device of Fig. 1.55 if the peak current is increased from 5 to 10 mA?
(b) Repeat part (a) for 30 to 35 mA (the same increase in current).
(c) Compare the percent increase from parts (a) and (b). At what point on the curve would you
say there is little gained by further increasing the peak current?
* 59. (a) Referring to Fig. 1.55h, determine the maximum tolerable peak current if the period of the
pulse duration is 1 ms, the frequency is 300 Hz, and the maximum tolerable dc current is
20 mA.
(b) Repeat part (a) for a frequency of 100 Hz.
60. (a) If the luminous intensity at 0° angular displacement is 3.0 mcd for the device of Fig. 1.55,
at what angle will it be 0.75 mcd?
(b) At what angle does the loss of luminous intensity drop below the 50% level?
* 61. Sketch the current derating curve for the average forward current of the high-efficiency red LED
of Fig. 1.55 as determined by temperature. (Note the absolute maximum ratings.)
*Please Note: Asterisks indicate more difficult problems.
50
Chapter 1
Semiconductor Diodes
CHAPTER
Diode
Applications
2
2.1 INTRODUCTION
The construction, characteristics, and models of semiconductor diodes were introduced in Chapter 1. The primary goal of this chapter is to develop a working knowledge of the diode in a variety of configurations using models appropriate for the area
of application. By chapter’s end, the fundamental behavior pattern of diodes in dc and
ac networks should be clearly understood. The concepts learned in this chapter will
have significant carryover in the chapters to follow. For instance, diodes are frequently
employed in the description of the basic construction of transistors and in the analysis of transistor networks in the dc and ac domains.
The content of this chapter will reveal an interesting and very positive side of the
study of a field such as electronic devices and systems—once the basic behavior of
a device is understood, its function and response in an infinite variety of configurations can be determined. The range of applications is endless, yet the characteristics
and models remain the same. The analysis will proceed from one that employs the
actual diode characteristic to one that utilizes the approximate models almost exclusively. It is important that the role and response of various elements of an electronic
system be understood without continually having to resort to lengthy mathematical
procedures. This is usually accomplished through the approximation process, which
can develop into an art itself. Although the results obtained using the actual characteristics may be slightly different from those obtained using a series of approximations, keep in mind that the characteristics obtained from a specification sheet may
in themselves be slightly different from the device in actual use. In other words, the
characteristics of a 1N4001 semiconductor diode may vary from one element to the
next in the same lot. The variation may be slight, but it will often be sufficient to validate the approximations employed in the analysis. Also consider the other elements
of the network: Is the resistor labeled 100 exactly 100 ? Is the applied voltage
exactly 10 V or perhaps 10.08 V? All these tolerances contribute to the general belief that a response determined through an appropriate set of approximations can often be “as accurate” as one that employs the full characteristics. In this book the emphasis is toward developing a working knowledge of a device through the use of
appropriate approximations, thereby avoiding an unnecessary level of mathematical
complexity. Sufficient detail will normally be provided, however, to permit a detailed
mathematical analysis if desired.
51
2.2 LOAD-LINE ANALYSIS
The applied load will normally have an important impact on the point or region of
operation of a device. If the analysis is performed in a graphical manner, a line can
be drawn on the characteristics of the device that represents the applied load. The intersection of the load line with the characteristics will determine the point of operation
of the system. Such an analysis is, for obvious reasons, called load-line analysis.
Although the majority of the diode networks analyzed in this chapter do not employ
the load-line approach, the technique is one used quite frequently in subsequent chapters, and this introduction offers the simplest application of the method. It also permits
a validation of the approximate technique described throughout the remainder of this
chapter.
Consider the network of Fig. 2.1a employing a diode having the characteristics
of Fig. 2.1b. Note in Fig. 2.1a that the “pressure” established by the battery is to establish a current through the series circuit in the clockwise direction. The fact that
this current and the defined direction of conduction of the diode are a “match” reveals that the diode is in the “on” state and conduction has been established. The resulting polarity across the diode will be as shown and the first quadrant (VD and ID
positive) of Fig. 2.1b will be the region of interest—the forward-bias region.
Applying Kirchhoff’s voltage law to the series circuit of Fig. 2.1a will result in
E VD VR 0
E VD ID R
or
(2.1)
The two variables of Eq. (2.1) (VD and ID) are the same as the diode axis variables of Fig. 2.1b. This similarity permits a plotting of Eq. (2.1) on the same characteristics of Fig. 2.1b.
The intersections of the load line on the characteristics can easily be determined
if one simply employs the fact that anywhere on the horizontal axis ID 0 A and
anywhere on the vertical axis VD 0 V.
If we set VD 0 V in Eq. (2.1) and solve for ID, we have the magnitude of ID on
the vertical axis. Therefore, with VD 0 V, Eq. (2.1) becomes
E VD IDR
0 V IDR
E
I D R
and
Figure 2.1 Series diode configuration: (a) circuit; (b) characteristics.
VD=0 V
(2.2)
as shown in Fig. 2.2. If we set ID 0 A in Eq. (2.1) and solve for VD, we have the
magnitude of VD on the horizontal axis. Therefore, with ID 0 A, Eq. (2.1) becomes
E VD ID R
VD (0 A)R
VD EID =0 A
and
(2.3)
as shown in Fig. 2.2. A straight line drawn between the two points will define the
load line as depicted in Fig. 2.2. Change the level of R (the load) and the intersection
on the vertical axis will change. The result will be a change in the slope of the load
line and a different point of intersection between the load line and the device characteristics.
We now have a load line defined by the network and a characteristic curve defined by the device. The point of intersection between the two is the point of opera52
Chapter 2
Diode Applications
ID
Characteristics (device)
E
R
Q-point
ID
Q
Load line (network)
0
VD
Q
E
VD
Figure 2.2 Drawing the load line and finding the point of operation.
tion for this circuit. By simply drawing a line down to the horizontal axis the diode
voltage VDQ can be determined, whereas a horizontal line from the point of intersection to the vertical axis will provide the level of IDQ. The current ID is actually the
current through the entire series configuration of Fig. 2.1a. The point of operation is
usually called the quiescent point (abbreviated “Q-pt.”) to reflect its “still, unmoving”
qualities as defined by a dc network.
The solution obtained at the intersection of the two curves is the same that would
be obtained by a simultaneous mathematical solution of Eqs. (2.1) and (1.4) [ID Is(ekVD/TK 1)]. Since the curve for a diode has nonlinear characteristics the mathematics involved would require the use of nonlinear techniques that are beyond the
needs and scope of this book. The load-line analysis described above provides a solution with a minimum of effort and a “pictorial” description of why the levels of solution for VDQ and IDQ were obtained. The next two examples will demonstrate the
techniques introduced above and reveal the relative ease with which the load line can
be drawn using Eqs. (2.2) and (2.3).
For the series diode configuration of Fig. 2.3a employing the diode characteristics of
Fig. 2.3b determine:
(a) VDQ and IDQ.
(b) VR.
EXAMPLE 2.1
Figure 2.3 (a) Circuit; (b) characteristics.
2.2 Load-line Analysis
53
Solution
E
10 V
(a) Eq. (2.2): ID 10 mA
R VD0 V 2 k
Eq. (2.3): VD EID0 A 10 V
The resulting load line appears in Fig. 2.4. The intersection between the load line and
the characteristic curve defines the Q-point as
VD 0.78 V
Q
ID 9.25 mA
Q
The level of VD is certainly an estimate, and the accuracy of ID is limited by the chosen scale. A higher degree of accuracy would require a plot that would be much larger
and perhaps unwieldy.
(b) VR IRR IDQR (9.25 mA)(1 k) 9.25 V
or VR E VD 10 V 0.78 V 9.22 V
The difference in results is due to the accuracy with which the graph can be read. Ideally, the results obtained either way should be the same.
Figure 2.4 Solution to Example 2.1.
EXAMPLE 2.2
Repeat the analysis of Example 2.1 with R 2 k.
Solution
E
10 V
(a) Eq. (2.2): ID 5 mA
R VD0 V 2 k
Eq. (2.3): VD EID0 A 10 V
The resulting load line appears in Fig. 2.5. Note the reduced slope and levels of diode
current for increasing loads. The resulting Q-point is defined by
VDQ 0.7 V
IDQ 4.6 mA
(b) VR IRR IDQR (4.6 mA)(2 k) 9.2 V
with VR E VD 10 V 0.7 V 9.3 V
The difference in levels is again due to the accuracy with which the graph can be read.
Certainly, however, the results provide an expected magnitude for the voltage VR.
54
Chapter 2
Diode Applications
I D (mA)
10
9
8
7
E
R 6
5
ID ~
= 4.6 mA
Q
4
3
2
1
Q-point
Load line
(from Example 2.1)
0.5 1
2
~ 0.7 V
VD =
Q
0
3
4
5
6
7
9
8
10 VD (V)
(E)
Figure 2.5 Solution to Example 2.2.
As noted in the examples above, the load line is determined solely by the applied
network while the characteristics are defined by the chosen device. If we turn to our
approximate model for the diode and do not disturb the network, the load line will
be exactly the same as obtained in the examples above. In fact, the next two examples repeat the analysis of Examples 2.1 and 2.2 using the approximate model to permit a comparison of the results.
Repeat Example 2.1 using the approximate equivalent model for the silicon semiconductor diode.
EXAMPLE 2.3
Solution
The load line is redrawn as shown in Fig. 2.6 with the same intersections as defined
in Example 2.1. The characteristics of the approximate equivalent circuit for the diode
have also been sketched on the same graph. The resulting Q-point:
VD 0.7 V
Q
ID 9.25 mA
Q
I D (mA)
10
ID ~
= 9.25 mA 9
Q
8
7
6
5
4
3
2
1
Q-point
0
0.5 1
2
~ 0.7 V
VD =
⇒
Load line
0.7 V
ID
3
4
5
6
7
8
9
10 VD (V)
Q
Figure 2.6 Solution to Example 2.1 using the diode approximate model.
2.2 Load-line Analysis
55
The results obtained in Example 2.3 are quite interesting. The level of IDQ is exactly the same as obtained in Example 2.1 using a characteristic curve that is a great
deal easier to draw than that appearing in Fig. 2.4. The level of VD 0.7 V versus
0.78 V from Example 2.1 is of a different magnitude to the hundredths place, but they
are certainly in the same neighborhood if we compare their magnitudes to the magnitudes of the other voltages of the network.
EXAMPLE 2.4
Repeat Example 2.2 using the approximate equivalent model for the silicon semiconductor diode.
Solution
The load line is redrawn as shown in Fig. 2.7 with the same intersections defined in
Example 2.2. The characteristics of the approximate equivalent circuit for the diode
have also been sketched on the same graph. The resulting Q-point:
VD 0.7 V
Q
ID 4.6 mA
I D (mA)
10
9
8
7
6
~
I D = 4.6 mA 5
Q
4
3
2
1
0
Q
⇒
0.7 V
ID
Q-point
Load line
0.5 1
2
~
VD = 0.7 V
3
4
5
6
7
9
8
10 VD (V)
Q
Figure 2.7 Solution to Example
2.2 using the diode approximate
model.
In Example 2.4 the results obtained for both VDQ and IDQ are the same as those
obtained using the full characteristics in Example 2.2. The examples above have
demonstrated that the current and voltage levels obtained using the approximate model
have been very close to those obtained using the full characteristics. It suggests, as
will be applied in the sections to follow, that the use of appropriate approximations
can result in solutions that are very close to the actual response with a reduced level
of concern about properly reproducing the characteristics and choosing a large-enough
scale. In the next example we go a step further and substitute the ideal model. The
results will reveal the conditions that must be satisfied to apply the ideal equivalent
properly.
EXAMPLE 2.4
Repeat Example 2.1 using the ideal diode model.
Solution
As shown in Fig. 2.8 the load line continues to be the same, but the ideal characteristics now intersect the load line on the vertical axis. The Q-point is therefore defined
by
VD 0 V
Q
ID 10 mA
Q
56
Chapter 2
Diode Applications
Figure 2.8 Solution to Example 2.1 using the ideal diode model.
The results are sufficiently different from the solutions of Example 2.1 to cause
some concern about their accuracy. Certainly, they do provide some indication of the
level of voltage and current to be expected relative to the other voltage levels of the
network, but the additional effort of simply including the 0.7-V offset suggests that
the approach of Example 2.3 is more appropriate.
Use of the ideal diode model therefore should be reserved for those occasions
when the role of a diode is more important than voltage levels that differ by tenths
of a volt and in those situations where the applied voltages are considerably larger
than the threshold voltage VT. In the next few sections the approximate model will be
employed exclusively since the voltage levels obtained will be sensitive to variations
that approach VT. In later sections the ideal model will be employed more frequently
since the applied voltages will frequently be quite a bit larger than VT and the authors
want to ensure that the role of the diode is correctly and clearly understood.
2.3 DIODE APPROXIMATIONS
In Section 2.2 we revealed that the results obtained using the approximate piecewiselinear equivalent model were quite close, if not equal, to the response obtained using
the full characteristics. In fact, if one considers all the variations possible due to tolerances, temperature, and so on, one could certainly consider one solution to be “as
accurate” as the other. Since the use of the approximate model normally results in a
reduced expenditure of time and effort to obtain the desired results, it is the approach
that will be employed in this book unless otherwise specified. Recall the following:
The primary purpose of this book is to develop a general knowledge of the behavior, capabilities, and possible areas of application of a device in a manner
that will minimize the need for extensive mathematical developments.
The complete piecewise-linear equivalent model introduced in Chapter 1 was not
employed in the load-line analysis because rav is typically much less than the other
series elements of the network. If rav should be close in magnitude to the other series
elements of the network, the complete equivalent model can be applied in much the
same manner as described in Section 2.2.
In preparation for the analysis to follow, Table 2.1 was developed to review the
important characteristics, models, and conditions of application for the approximate
and ideal diode models. Although the silicon diode is used almost exclusively due to
2.3 Diode Approximations
57
TABLE 2.1 Approximate and Ideal Semiconductor Diode Models
its temperature characteristics, the germanium diode is still employed and is therefore included in Table 2.1. As with the silicon diode, a germanium diode is approximated by an open-circuit equivalent for voltages less than VT. It will enter the “on”
state when VD VT 0.3 V.
Keep in mind that the 0.7 and 0.3 V in the equivalent circuits are not independent
sources of energy but are there simply to remind us that there is a “price to pay” to
turn on a diode. An isolated diode on a laboratory table will not indicate 0.7 or 0.3
V if a voltmeter is placed across its terminals. The supplies specify the voltage drop
across each when the device is “on” and specify that the diode voltage must be at
least the indicated level before conduction can be established.
58
Chapter 2
Diode Applications
In the next few sections we demonstrate the impact of the models of Table 2.1 on
the analysis of diode configurations. For those situations where the approximate equivalent circuit will be employed, the diode symbol will appear as shown in Fig. 2.9a
for the silicon and germanium diodes. If conditions are such that the ideal diode model
can be employed, the diode symbol will appear as shown in Fig. 2.9b.
2.4 SERIES DIODE CONFIGURATIONS
WITH DC INPUTS
Figure 2.9 (a) Approximate
model notation; (b) ideal diode
notation.
In this section the approximate model is utilized to investigate a number of series
diode configurations with dc inputs. The content will establish a foundation in diode
analysis that will carry over into the sections and chapters to follow. The procedure
described can, in fact, be applied to networks with any number of diodes in a variety
of configurations.
For each configuration the state of each diode must first be determined. Which
diodes are “on” and which are “off”? Once determined, the appropriate equivalent as
defined in Section 2.3 can be substituted and the remaining parameters of the network determined.
In general, a diode is in the “on” state if the current established by the
applied sources is such that its direction matches that of the arrow in the
diode symbol, and VD 0.7 V for silicon and VD 0.3 V for germanium.
For each configuration, mentally replace the diodes with resistive elements and
note the resulting current direction as established by the applied voltages (“pressure”).
If the resulting direction is a “match” with the arrow in the diode symbol, conduction through the diode will occur and the device is in the “on” state. The description
above is, of course, contingent on the supply having a voltage greater than the “turnon” voltage (VT) of each diode.
If a diode is in the “on” state, one can either place a 0.7-V drop across the
element, or the network can be redrawn with the VT equivalent circuit as defined in
Table 2.1. In time the preference will probably simply be to include the 0.7-V drop across
each “on” diode and draw a line through each diode in the “off” or open state. Initially, however, the substitution method will be utilized to ensure that the proper voltage and current levels are determined.
The series circuit of Fig. 2.10 described in some detail in Section 2.2 will be used
to demonstrate the approach described in the paragraphs above. The state of the diode
is first determined by mentally replacing the diode with a resistive element as shown
in Fig. 2.11. The resulting direction of I is a match with the arrow in the diode symbol, and since E VT the diode is in the “on” state. The network is then redrawn as
shown in Fig. 2.12 with the appropriate equivalent model for the forward-biased silicon diode. Note for future reference that the polarity of VD is the same as would result if in fact the diode were a resistive element. The resulting voltage and current
levels are the following:
VD VT
(2.4)
VR E VT
(2.5)
VR
ID IR R
(2.6)
2.4 Series Diode Configurations with DC Inputs
Figure 2.10 Series diode configuration.
+
E
I
+
R
–
VR
–
Figure 2.11 Determining the
state of the diode of Fig. 2.10.
Figure 2.12 Substituting the
equivalent model for the “on”
diode of Fig. 2.10.
59
Figure 2.14 Determining the
state of the diode of Fig. 2.13.
Figure 2.13 Reversing the diode
of Fig. 2.10.
Figure 2.15 Substituting the
equivalent model for the “off”
diode of Figure 2.13.
In Fig. 2.13 the diode of Fig. 2.10 has been reversed. Mentally replacing the diode
with a resistive element as shown in Fig. 2.14 will reveal that the resulting current direction does not match the arrow in the diode symbol. The diode is in the “off” state,
resulting in the equivalent circuit of Fig. 2.15. Due to the open circuit, the diode current is 0 A and the voltage across the resistor R is the following:
VR IRR ID R (0 A)R 0 V
The fact that VR 0 V will establish E volts across the open circuit as defined by
Kirchhoff’s voltage law. Always keep in mind that under any circumstances—dc, ac
instantaneous values, pulses, and so on—Kirchhoff’s voltage law must be satisfied!
EXAMPLE 2.6
For the series diode configuration of Fig. 2.16, determine VD, VR, and ID.
Solution
Since the applied voltage establishes a current in the clockwise direction to match the
arrow of the symbol and the diode is in the “on” state,
VD 0.7 V
VR E VD 8 V 0.7 V 7.3 V
7.3 V
VR
ID IR 3.32 mA
2.2 k
R
Figure 2.16 Circuit for Example
2.6.
Repeat Example 2.6 with the diode reversed.
EXAMPLE 2.7
ID = 0 A
Solution
+ VD –
IR = 0 A
+
E
8V
R
2.2 kΩ VR
–
Removing the diode, we find that the direction of I is opposite to the arrow in the
diode symbol and the diode equivalent is the open circuit no matter which model is
employed. The result is the network of Fig. 2.17, where ID 0 A due to the open circuit. Since VR IRR, VR (0)R 0 V. Applying Kirchhoff’s voltage law around the
closed loop yields
E VD VR 0
Figure 2.17 Determining the
unknown quantities for Example
2.7.
and
60
Chapter 2
VD E VR E 0 E 8 V
Diode Applications
In particular, note in Example 2.7 the high voltage across the diode even though
it is an “off” state. The current is zero, but the voltage is significant. For review purposes, keep the following in mind for the analysis to follow:
1. An open circuit can have any voltage across its terminals, but the current is always 0 A.
2. A short circuit has a 0-V drop across its terminals, but the current is limited
only by the surrounding network.
In the next example the notation of Fig. 2.18 will be employed for the applied voltage. It is a common industry notation and one with which the reader should become very
familiar. Such notation and other defined voltage levels are treated further in Chapter 4.
E = + 10 V
+10 V
E
–5 V
E = –5 V
10 V
E
5V
Figure 2.18 Source notation.
For the series diode configuration of Fig. 2.19, determine VD, VR, and ID.
EXAMPLE 2.8
Figure 2.19 Series diode circuit
for Example 2.8.
Solution
Although the “pressure” establishes a current with the same direction as the arrow
symbol, the level of applied voltage is insufficient to turn the silicon diode “on.” The
point of operation on the characteristics is shown in Fig. 2.20, establishing the opencircuit equivalent as the appropriate approximation. The resulting voltage and current
levels are therefore the following:
ID 0 A
VR IRR ID R (0 A)1.2 k 0 V
and
VD E 0.5 V
Figure 2.20 Operating point
with E 0.5 V.
2.4 Series Diode Configurations with DC Inputs
61
EXAMPLE 2.9
Determine Vo and ID for the series circuit of Fig. 2.21.
Figure 2.21 Circuit for Example 2.9.
Solution
An attack similar to that applied in Example 2.6 will reveal that the resulting current
has the same direction as the arrowheads of the symbols of both diodes, and the network of Fig. 2.22 results because E 12 V (0.7 V 0.3 V) 1 V. Note the redrawn supply of 12 V and the polarity of Vo across the 5.6-k resistor. The resulting
voltage
Vo E VT VT 12 V 0.7 V 0.3 V 11 V
1
2
VR
V
11 V
ID IR o 1.96 mA
R
R
5.6 k
and
Figure 2.22 Determining the
unknown quantities for Example
2.9.
EXAMPLE 2.10
Determine ID, VD2, and Vo for the circuit of Fig. 2.23.
+
Si
+12 V
VD
2
–
Si
IR
ID
Vo
5.6 kΩ
Figure 2.23 Circuit for Example 2.10.
Solution
Removing the diodes and determining the direction of the resulting current I will result in the circuit of Fig. 2.24. There is a match in current direction for the silicon
diode but not for the germanium diode. The combination of a short circuit in series
with an open circuit always results in an open circuit and ID 0 A, as shown in
Fig. 2.25.
+
I
E
R
5.6 kΩ Vo
–
Figure 2.24 Determining the state of the
diodes of Figure 2.23.
62
Chapter 2
Diode Applications
Figure 2.25 Substituting the equivalent
state for the open diode.
The question remains as to what to substitute for the silicon diode. For the analysis to follow in this and succeeding chapters, simply recall for the actual practical
diode that when ID 0 A, VD 0 V (and vice versa), as described for the no-bias
situation in Chapter 1. The conditions described by ID 0 A and VD1 0 V are indicated in Fig. 2.26.
Figure 2.26 Determining the
unknown quantities for the circuit
of Example 2.10.
Vo IRR IDR (0 A)R 0 V
and
VD Vopen circuit E 12 V
2
Applying Kirchhoff’s voltage law in a clockwise direction gives us
E VD VD Vo 0
1
and
2
VD E VD Vo 12 V 0 0
2
1
12 V
with
Vo 0 V
Determine I, V1, V2, and Vo for the series dc configuration of Fig. 2.27.
EXAMPLE 2.11
Figure 2.27 Circuit for Example 2.11.
Solution
The sources are drawn and the current direction indicated as shown in Fig. 2.28. The
diode is in the “on” state and the notation appearing in Fig. 2.29 is included to indicate this state. Note that the “on” state is noted simply by the additional VD 0.7 V
Figure 2.28 Determining the state of the
diode for the network of Fig. 2.27.
Figure 2.29 Determining the unknown quantities for the network of Fig. 2.27.
2.4 Series Diode Configurations with DC Inputs
63
on the figure. This eliminates the need to redraw the network and avoids any confusion that may result from the appearance of another source. As indicated in the introduction to this section, this is probably the path and notation that one will take
when a level of confidence has been established in the analysis of diode configurations. In time the entire analysis will be performed simply by referring to the original network. Recall that a reverse-biased diode can simply be indicated by a line
through the device.
The resulting current through the circuit is,
10 V 5 V 0.7 V
E1 E2 VD
14.3 V
I
R1 R2
4.7 k 2.2 k
6.9 k
2.072 mA
and the voltages are
V1 IR1 (2.072 mA)(4.7 k) 9.74 V
V2 IR2 (2.072 mA)(2.2 k) 4.56 V
Applying Kirchhoff’s voltage law to the output section in the clockwise direction will
result in
E2 V2 Vo 0
and
Vo V2 E2 4.56 V 5 V 0.44 V
The minus sign indicates that Vo has a polarity opposite to that appearing in Fig. 2.27.
2.5 PARALLEL AND SERIES–PARALLEL
CONFIGURATIONS
The methods applied in Section 2.4 can be extended to the analysis of parallel and
series–parallel configurations. For each area of application, simply match the sequential series of steps applied to series diode configurations.
EXAMPLE 2.12
Determine Vo, I1, ID1, and ID2 for the parallel diode configuration of Fig. 2.30.
Figure 2.30 Network for Example 2.12.
Solution
For the applied voltage the “pressure” of the source is to establish a current through
each diode in the same direction as shown in Fig. 2.31. Since the resulting current direction matches that of the arrow in each diode symbol and the applied voltage is
greater than 0.7 V, both diodes are in the “on” state. The voltage across parallel elements is always the same and
Vo 0.7 V
64
Chapter 2
Diode Applications
Figure 2.31 Determining the
unknown quantities for the network of Example 2.12.
The current
VR
E VD
10 V 0.7 V
I1 28.18 mA
R
0.33 k
R
Assuming diodes of similar characteristics, we have
I
28.18 mA
ID ID 1 14.09 mA
2
2
1
2
Example 2.12 demonstrated one reason for placing diodes in parallel. If the current rating of the diodes of Fig. 2.30 is only 20 mA, a current of 28.18 mA would
damage the device if it appeared alone in Fig. 2.30. By placing two in parallel, the
current is limited to a safe value of 14.09 mA with the same terminal voltage.
EXAMPLE 2.13
Determine the current I for the network of Fig. 2.32.
Figure 2.32 Network for Example 2.13.
Solution
Redrawing the network as shown in Fig. 2.33 reveals that the resulting current direction is such as to turn on diode D1 and turn off diode D2. The resulting current I
is then
E1 E2 VD
20 V 4 V 0.7 V
6.95 mA
I
2.2 k
R
Figure 2.33 Determining the
unknown quantities for the network of Example 2.13.
2.5 Parallel and Series–Parallel Configurations
65
EXAMPLE 2.14
Determine the voltage Vo for the network of Fig. 2.34.
Solution
12 V
Si
Ge
Vo
2.2 kΩ
Initially, it would appear that the applied voltage will turn both diodes “on.” However,
if both were “on,” the 0.7-V drop across the silicon diode would not match the 0.3 V
across the germanium diode as required by the fact that the voltage across parallel elements must be the same. The resulting action can be explained simply by realizing
that when the supply is turned on it will increase from 0 to 12 V over a period of
time—although probably measurable in milliseconds. At the instant during the rise
that 0.3 V is established across the germanium diode it will turn “on” and maintain
a level of 0.3 V. The silicon diode will never have the opportunity to capture its required 0.7 V and therefore remains in its open-circuit state as shown in Fig. 2.35. The
result:
Vo 12 V 0.3 V 11.7 V
Figure 2.34
ple 2.14.
Network for Exam-
Figure 2.35 Determining Vo
for the network of Fig. 2.34.
EXAMPLE 2.15
Determine the currents I1, I2, and ID2 for the network of Fig. 2.36.
Solution
The applied voltage (pressure) is such as to turn both diodes on, as noted by the resulting current directions in the network of Fig. 2.37. Note the use of the abbreviated
notation for “on” diodes and that the solution is obtained through an application of
techniques applied to dc series—parallel networks.
0.7 V
V
I1 T 0.212 mA
R1
3.3 k
2
Figure 2.36 Network for Example 2.15.
Figure 2.37 Determining the
unknown quantities for Example
2.15.
66
Chapter 2
Diode Applications
Applying Kirchhoff’s voltage law around the indicated loop in the clockwise direction yields
V2 E VT VT 0
1
2
and
V2 E VT VT 20 V 0.7 V 0.7 V 18.6 V
with
V2
18.6 V
I2 3.32 mA
R2
5.6 k
1
2
At the bottom node (a),
ID2 I1 I2
and
ID2 I2 I1 3.32 mA 0.212 mA 3.108 mA
2.6 AND/OR GATES
The tools of analysis are now at our disposal, and the opportunity to investigate a
computer configuration is one that will demonstrate the range of applications of this
relatively simple device. Our analysis will be limited to determining the voltage levels and will not include a detailed discussion of Boolean algebra or positive and negative logic.
The network to be analyzed in Example 2.16 is an OR gate for positive logic.
That is, the 10-V level of Fig. 2.38 is assigned a “1” for Boolean algebra while the
0-V input is assigned a “0.” An OR gate is such that the output voltage level will be
a 1 if either or both inputs is a 1. The output is a 0 if both inputs are at the 0 level.
The analysis of AND/OR gates is made measurably easier by using the approximate equivalent for a diode rather than the ideal because we can stipulate that the
voltage across the diode must be 0.7 V positive for the silicon diode (0.3 V for Ge)
to switch to the “on” state.
In general, the best approach is simply to establish a “gut” feeling for the state of
the diodes by noting the direction and the “pressure” established by the applied potentials. The analysis will then verify or negate your initial assumptions.
Figure 2.38 Positive logic OR
gate.
EXAMPLE 2.16
Determine Vo for the network of Fig. 2.38.
Solution
First note that there is only one applied potential; 10 V at terminal 1. Terminal 2 with
a 0-V input is essentially at ground potential, as shown in the redrawn network of Fig.
2.39. Figure 2.39 “suggests” that D1 is probably in the “on” state due to the applied
10 V while D2 with its “positive” side at 0 V is probably “off.” Assuming these states
will result in the configuration of Fig. 2.40.
The next step is simply to check that there is no contradiction to our assumptions.
That is, note that the polarity across D1 is such as to turn it on and the polarity across
D2 is such as to turn it off. For D1 the “on” state establishes Vo at Vo E VD 10 V 0.7 V 9.3 V. With 9.3 V at the cathode () side of D2 and 0 V at the anode () side, D2 is definitely in the “off” state. The current direction and the resulting continuous path for conduction further confirm our assumption that D1 is conducting. Our assumptions seem confirmed by the resulting voltages and current, and
our initial analysis can be assumed to be correct. The output voltage level is not 10
V as defined for an input of 1, but the 9.3 V is sufficiently large to be considered a
1 level. The output is therefore at a 1 level with only one input, which suggests that
2.6 And/Or Gates
+
–
D1
E
Vo
10 V
D2
R
1 kΩ
0V
Figure 2.39 Redrawn network
of Fig. 2.38.
67
Figure 2.40 Assumed diode
states for Fig. 2.38.
the gate is an OR gate. An analysis of the same network with two 10-V inputs will
result in both diodes being in the “on” state and an output of 9.3 V. A 0-V input at
both inputs will not provide the 0.7 V required to turn the diodes on, and the output
will be a 0 due to the 0-V output level. For the network of Fig. 2.40 the current level
is determined by
E VD
10 V 0.7 V
9.3 mA
I R
1 k
EXAMPLE 2.17
(1)
E1 = 10 V
(0)
E2 = 0 V
Determine the output level for the positive logic AND gate of Fig. 2.41.
Solution
Si
1
D1
Si
2
Vo
D2
R
E
Figure 2.41
gate.
1 kΩ
10 V
Positive logic AND
Note in this case that an independent source appears in the grounded leg of the network. For reasons soon to become obvious it is chosen at the same level as the input
logic level. The network is redrawn in Fig. 2.42 with our initial assumptions regarding the state of the diodes. With 10 V at the cathode side of D1 it is assumed that D1
is in the “off” state even though there is a 10-V source connected to the anode of D1
through the resistor. However, recall that we mentioned in the introduction to this section that the use of the approximate model will be an aid to the analysis. For D1,
where will the 0.7 V come from if the input and source voltages are at the same level
and creating opposing “pressures”? D2 is assumed to be in the “on” state due to the
low voltage at the cathode side and the availability of the 10-V source through the
1-k resistor.
For the network of Fig. 2.42 the voltage at Vo is 0.7 V due to the forward-biased
diode D2. With 0.7 V at the anode of D1 and 10 V at the cathode, D1 is definitely in
the “off” state. The current I will have the direction indicated in Fig. 2.42 and a magnitude equal to
E VD
10 V 0.7 V
9.3 mA
I R
1 k
VD
–
(1)
E1
+
Vo = VD = 0.7 V (0)
0.7V
10 V
R
1 kΩ
(0)
I
68
Chapter 2
E
Diode Applications
10 V
Figure 2.42 Substituting the
assumed states for the diodes of
Fig. 2.41.
The state of the diodes is therefore confirmed and our earlier analysis was correct. Although not 0 V as earlier defined for the 0 level, the output voltage is sufficiently small to be considered a 0 level. For the AND gate, therefore, a single input
will result in a 0-level output. The remaining states of the diodes for the possibilities
of two inputs and no inputs will be examined in the problems at the end of the
chapter.
2.7 SINUSOIDAL INPUTS; HALF-WAVE
RECTIFICATION
The diode analysis will now be expanded to include time-varying functions such as
the sinusoidal waveform and the square wave. There is no question that the degree of
difficulty will increase, but once a few fundamental maneuvers are understood, the
analysis will be fairly direct and follow a common thread.
The simplest of networks to examine with a time-varying signal appears in Fig.
2.43. For the moment we will use the ideal model (note the absence of the Si or Ge
label to denote ideal diode) to ensure that the approach is not clouded by additional
mathematical complexity.
vi
+
+
T
2
0
–
+
Vm
T
vi
t
vo
R
–
–
1 cycle
vi = Vm sin ωt
Figure 2.43
Half-wave rectifier.
Over one full cycle, defined by the period T of Fig. 2.43, the average value (the
algebraic sum of the areas above and below the axis) is zero. The circuit of Fig. 2.43,
called a half-wave rectifier, will generate a waveform vo that will have an average
value of particular, use in the ac-to-dc conversion process. When employed in the rectification process, a diode is typically referred to as a rectifier. Its power and current
ratings are typically much higher than those of diodes employed in other applications,
such as computers and communication systems.
During the interval t 0 → T/2 in Fig. 2.43 the polarity of the applied voltage vi
is such as to establish “pressure” in the direction indicated and turn on the diode with
the polarity appearing above the diode. Substituting the short-circuit equivalence for
the ideal diode will result in the equivalent circuit of Fig. 2.44, where it is fairly obvious that the output signal is an exact replica of the applied signal. The two terminals defining the output voltage are connected directly to the applied signal via the
short-circuit equivalence of the diode.
+
–
+
vi
–
Figure 2.44
R
+
+
vo
vi
–
–
+
R
vo
Vm
vo = vi
–
0
T
2
t
Conduction region (0 → T/2).
2.7 Sinusoidal Inputs; Half-Wave Rectification
69
For the period T/2 → T, the polarity of the input vi is as shown in Fig. 2.45 and
the resulting polarity across the ideal diode produces an “off” state with an open-circuit equivalent. The result is the absence of a path for charge to flow and vo iR (0)R 0 V for the period T/2 → T. The input vi and the output vo were sketched together in Fig. 2.46 for comparison purposes. The output signal vo now has a net positive area above the axis over a full period and an average value determined by
Vdc 0.318Vm
–
(2.7)
half-wave
+
–
vi
R
+
Figure 2.45
+
–
vo
vi
–
+
+
vo
vo = 0 V
vo = 0 V
R
0
–
T
2
T
t
Nonconduction region (T/2 → T).
vi
Vm
t
0
Vdc = 0 V
vo
Vm
Vdc = 0.318Vm
0
t
T
Figure 2.46
signal.
Half-wave rectified
The process of removing one-half the input signal to establish a dc level is aptly
called half-wave rectification.
The effect of using a silicon diode with VT 0.7 V is demonstrated in Fig. 2.47
for the forward-bias region. The applied signal must now be at least 0.7 V before the
diode can turn “on.” For levels of vi less than 0.7 V, the diode is still in an opencircuit state and vo 0 V as shown in the same figure. When conducting, the difference between vo and vi is a fixed level of VT 0.7 V and vo vi VT , as shown in
the figure. The net effect is a reduction in area above the axis, which naturally reduces
vi
+
Vm
0
+
T
2
VT = 0.7 V
t
T
vi
VT
–
vo
+
0.7 V
R
vo
0
–
Vm – VT
–
T
2
Offset due to VT
Figure 2.47
70
Chapter 2
Effect of VT on half-wave rectified signal.
Diode Applications
Tt
the resulting dc voltage level. For situations where Vm VT, Eq. 2.8 can be applied
to determine the average value with a relatively high level of accuracy.
Vdc 0.318(Vm VT)
(2.8)
In fact, if Vm is sufficiently greater than VT, Eq. 2.7 is often applied as a first approximation for Vdc.
EXAMPLE 2.18
(a) Sketch the output vo and determine the dc level of the output for the network of
Fig. 2.48.
(b) Repeat part (a) if the ideal diode is replaced by a silicon diode.
(c) Repeat parts (a) and (b) if Vm is increased to 200 V and compare solutions using
Eqs. (2.7) and (2.8).
+
vi
20 V
vi
T t
T
2
0
+
R
vo
2 kΩ
–
–
Figure 2.48
ple 2.18.
Network for Exam-
Solution
(a) In this situation the diode will conduct during the negative part of the input as
shown in Fig. 2.49, and vo will appear as shown in the same figure. For the full
period, the dc level is
Vdc 0.318Vm 0.318(20 V) 6.36 V
The negative sign indicates that the polarity of the output is opposite to the defined
polarity of Fig. 2.48.
–
vi
T
2
T
20
Figure 2.49
vo
–
20
0
+
t
vi
+
2 kΩ
vo
–
+
0
T
2
T
t
20 V
Resulting vo for the circuit of Example 2.18.
(b) Using a silicon diode, the output has the appearance of Fig. 2.50 and
Vdc 0.318(Vm 0.7 V) 0.318(19.3 V) 6.14 V
The resulting drop in dc level is 0.22 V or about 3.5%.
(c) Eq. (2.7): Vdc 0.318Vm 0.318(200 V) 63.6 V
Eq. (2.8): Vdc 0.318(Vm VT) 0.318(200 V 0.7 V)
(0.318)(199.3 V) 63.38 V
which is a difference that can certainly be ignored for most applications. For part c
the offset and drop in amplitude due to VT would not be discernible on a typical oscilloscope if the full pattern is displayed.
2.7 Sinusoidal Inputs; Half-Wave Rectification
vo
0
T
2
T
t
20 V – 0.7 V = 19.3 V
Figure 2.50 Effect of VT on output of Fig. 2.49.
71
PIV (PRV)
The peak inverse voltage (PIV) [or PRV (peak reverse voltage)] rating of the diode
is of primary importance in the design of rectification systems. Recall that it is the
voltage rating that must not be exceeded in the reverse-bias region or the diode will
enter the Zener avalanche region. The required PIV rating for the half-wave rectifier
can be determined from Fig. 2.51, which displays the reverse-biased diode of Fig.
2.43 with maximum applied voltage. Applying Kirchhoff”s voltage law, it is fairly
obvious that the PIV rating of the diode must equal or exceed the peak value of the
applied voltage. Therefore,
PIV rating
–
V (PIV)
–
(2.9)
half-wave rectifier
+
–
I= 0
Vm
Vm
Vo = IR = (0)R = 0 V
R
+
+
Figure 2.51 Determining the required PIV rating for the halfwave rectifier.
2.8 FULL-WAVE RECTIFICATION
Bridge Network
The dc level obtained from a sinusoidal input can be improved 100% using a process
called full-wave rectification. The most familiar network for performing such a function appears in Fig. 2.52 with its four diodes in a bridge configuration. During the
period t 0 to T/2 the polarity of the input is as shown in Fig. 2.53. The resulting
polarities across the ideal diodes are also shown in Fig. 2.53 to reveal that D2 and D3
are conducting while D1 and D4 are in the “off” state. The net result is the configuration of Fig. 2.54, with its indicated current and polarity across R. Since the diodes
are ideal the load voltage is vo vi, as shown in the same figure.
vi
+
D1
Vm
T
2
0
T
–
vi
t
"off "
–
vi
+
–
–
vo
+
–
+
D3
D4
Figure 2.52 Full-wave
bridge rectifier.
"on"
vi
–
vo
+
Vm
R
+
"on"
+
vo
R
–
+
D2
Vm
+
–
"off "
R
vi
0
T
2
–
t
vo
+
–
Figure 2.53 Network of Fig.
2.52 for the period 0 → T/2 of
the input voltage vi.
Figure 2.54
72
Chapter 2
Conduction path for the positive region of vi.
Diode Applications
0
T
2
t
For the negative region of the input the conducting diodes are D1 and D4, resulting in the configuration of Fig. 2.55. The important result is that the polarity across
the load resistor R is the same as in Fig. 2.53, establishing a second positive pulse,
as shown in Fig. 2.55. Over one full cycle the input and output voltages will appear
as shown in Fig. 2.56.
vi
vo
–
Vm
0
T
T
2
–
vi
vo
t
+
0
R
T
2
T
t
Vm
+
Figure 2.55
Conduction path for the negative region of vi.
vo
vi
Vm
Vm
Vdc = 0.636Vm
0
T
T
2
T
2
0
t
T t
Figure 2.56 Input and output
waveforms for a full-wave rectifier.
Since the area above the axis for one full cycle is now twice that obtained for a
half-wave system, the dc level has also been doubled and
Vdc 2(Eq. 2.7) 2(0.318Vm)
Vdc 0.636Vm
or
full-wave
(2.10)
If silicon rather than ideal diodes are employed as shown in Fig. 2.57, an application of Kirchhoff’s voltage law around the conduction path would result in
vi VT vo VT 0
vo vi 2VT
and
The peak value of the output voltage vo is therefore
Vomax Vm 2VT
For situations where Vm 2VT, Eq. (2.11) can be applied for the average value with
a relatively high level of accuracy.
Vdc 0.636(Vm 2VT)
+
+
vo
VT = 0.7 V
–
–
vi
–
Vm – 2VT
+
R
VT = 0.7 V
+
–
vo
(2.11)
0
T
2
T
t
Figure 2.57 Determining Vomax for
silicon diodes in the bridge configuration.
Then again, if Vm is sufficiently greater than 2VT, then Eq. (2.10) is often applied as
a first approximation for Vdc.
2.8 Full-Wave Rectification
73
PIV
The required PIV of each diode (ideal) can be determined from Fig. 2.58 obtained
at the peak of the positive region of the input signal. For the indicated loop the maximum voltage across R is Vm and the PIV rating is defined by
PIV
Vm
(2.12)
full-wave bridge rectifier
Center-Tapped Transformer
Figure 2.58 Determining the required PIV for the bridge configuration.
A second popular full-wave rectifier appears in Fig. 2.59 with only two diodes but
requiring a center-tapped (CT) transformer to establish the input signal across each
section of the secondary of the transformer. During the positive portion of vi applied
to the primary of the transformer, the network will appear as shown in Fig. 2.60. D1
assumes the short-circuit equivalent and D2 the open-circuit equivalent, as determined
by the secondary voltages and the resulting current directions. The output voltage appears as shown in Fig. 2.60.
D1
1:2
+
vi
vi
Vm
+
t
0
R
–
vi
CT
–
+
–
vo
+
vi
–
Figure 2.59 Center-tapped
transformer full-wave rectifier.
D2
1:2
vi
+
vo
Vm
Vm
Vm
+
0
T
2
t
vi
–
+
–
Figure 2.60
–
–
CT
vo
+
0
R
Vm
T
2
t
– +
Network conditions for the positive region of vi.
During the negative portion of the input the network appears as shown in Fig.
2.61, reversing the roles of the diodes but maintaining the same polarity for the voltvi
–
vo
– +
Vm
–
0
T
2
T
Vm
t
vi
+
+
CT
74
Chapter 2
R
–
–
+
Figure 2.61
Vm
vo
+
Vm
Network conditions for the negative region of vi.
Diode Applications
0
T
2
T
t
age across the load resistor R. The net effect is the same output as that appearing in
Fig. 2.56 with the same dc levels.
PIV
The network of Fig. 2.62 will help us determine the net PIV for each diode for
this full-wave rectifier. Inserting the maximum voltage for the secondary voltage and
Vm as established by the adjoining loop will result in
PIV Vsecondary VR
Vm Vm
and
PIV
2Vm
Figure 2.62 Determining the
PIV level for the diodes of the CT
transformer full-wave rectifier.
(2.13)
CT transformer, full-wave rectifier
EXAMPLE 2.19
Determine the output waveform for the network of Fig. 2.63 and calculate the output
dc level and the required PIV of each diode.
vi
+
10 V
2 kΩ
vi
T
2
0
–
T t
+
vo
2 kΩ
2 kΩ
–
Figure 2.63 Bridge network for
Example 2.19.
Solution
The network will appear as shown in Fig. 2.64 for the positive region of the input
voltage. Redrawing the network will result in the configuration of Fig. 2.65, where
vo 12 vi or Vomax 12 Vi max 12 (10 V) 5 V, as shown in Fig. 2.65. For the negative
part of the input the roles of the diodes will be interchanged and vo will appear as
shown in Fig. 2.66.
+
vi
+
+
2 kΩ
–
10 V
2 kΩ
vi
0
T
2
–
Figure 2.64
region of vi.
–
t
2 kΩ
+
vo
vo
–
vi
+
2 kΩ
2 kΩ
Network of Fig. 2.63 for the positive
vo
5V
2 kΩ
0
T
2
t
–
Figure 2.65
Redrawn network of Fig. 2.64.
The effect of removing two diodes from the bridge configuration was therefore to
reduce the available dc level to the following:
vo
5V
Vdc 0.636(5 V) 3.18 V
or that available from a half-wave rectifier with the same input. However, the PIV as
determined from Fig. 2.58 is equal to the maximum voltage across R, which is 5 V
or half of that required for a half-wave rectifier with the same input.
2.8 Full-Wave Rectification
0
T
2
T
t
Figure 2.66 Resulting output
for Example 2.19.
75
2.9 CLIPPERS
There are a variety of diode networks called clippers that have the ability to “clip”
off a portion of the input signal without distorting the remaining part of the alternating waveform. The half-wave rectifier of Section 2.7 is an example of the simplest
form of diode clipper—one resistor and diode. Depending on the orientation of the
diode, the positive or negative region of the input signal is “clipped” off.
There are two general categories of clippers: series and parallel. The series configuration is defined as one where the diode is in series with the load, while the parallel variety has the diode in a branch parallel to the load.
Series
The response of the series configuration of Fig. 2.67a to a variety of alternating waveforms is provided in Fig. 2.67b. Although first introduced as a half-wave rectifier (for
sinusoidal waveforms), there are no boundaries on the type of signals that can be applied to a clipper. The addition of a dc supply such as shown in Fig. 2.68 can have a
pronounced effect on the output of a clipper. Our initial discussion will be limited to
ideal diodes, with the effect of VT reserved for a concluding example.
+
+
vi
R
vo
–
–
(a)
vi
vo
V
0
V
t
vo
vi
V
V
t
–V
t
t
–V
(b)
Figure 2.67
Series clipper.
Figure 2.68
a dc supply.
Series clipper with
There is no general procedure for analyzing networks such as the type in Fig.
2.68, but there are a few thoughts to keep in mind as you work toward a solution.
1. Make a mental sketch of the response of the network based on the direction of the diode and the applied voltage levels.
For the network of Fig. 2.68, the direction of the diode suggests that the signal vi
must be positive to turn it on. The dc supply further requires that the voltage vi be
greater than V volts to turn the diode on. The negative region of the input signal is
76
Chapter 2
Diode Applications
“pressuring” the diode into the “off” state, supported further by the dc supply. In general, therefore, we can be quite sure that the diode is an open circuit (“off” state) for
the negative region of the input signal.
2. Determine the applied voltage (transition voltage) that will cause a change
in state for the diode.
For the ideal diode the transition between states will occur at the point on the
characteristics where vd 0 V and id 0 A. Applying the condition id 0 at vd 0 to the network of Fig. 2.68 will result in the configuration of Fig. 2.69, where it is
recognized that the level of vi that will cause a transition in state is
vi V
V
vd = 0 V
+
–
(2.14)
id = 0 A
+
+
vi
R
–
vo = iRR = id R = (0)R = 0 V
Figure 2.69 Determining the
transition level for the circuit of
Fig. 2.68.
–
For an input voltage greater than V volts the diode is in the short-circuit state, while
for input voltages less than V volts it is in the open-circuit or “off” state.
3. Be continually aware of the defined terminals and polarity of vo.
When the diode is in the short-circuit state, such as shown in Fig. 2.70, the output voltage vo can be determined by applying Kirchhoff’s voltage law in the clockwise direction:
Figure 2.70
Determining vo.
Figure 2.71
levels of vo.
Determining
vi V vo 0 (CW direction)
vo vi V
and
(2.15)
4. It can be helpful to sketch the input signal above the output and determine
the output at instantaneous values of the input.
It is then possible that the output voltage can be sketched from the resulting data
points of vo as demonstrated in Fig. 2.71. Keep in mind that at an instantaneous value
of vi the input can be treated as a dc supply of that value and the corresponding dc
value (the instantaneous value) of the output determined. For instance, at vi Vm
for the network of Fig. 2.68, the network to be analyzed appears in Fig. 2.72. For Vm
V the diode is in the short-circuit state and vo Vm V, as shown in Fig. 2.71.
At vi V the diodes change state; at vi Vm, vo 0 V; and the complete curve
for vo can be sketched as shown in Fig. 2.73.
vi
vo
Vm
Vm – V
V
0
T
2
T
t
0
T
2
T
t
vi = V (diodes change state)
Figure 2.72
Determining vo when vi Vm.
Figure 2.73
Sketching vo.
2.9 Clippers
77
EXAMPLE 2.20
Determine the output waveform for the network of Fig. 2.74.
Figure 2.74 Series clipper for
Example 2.20.
Solution
Past experience suggests that the diode will be in the “on” state for the positive region of vi —especially when we note the aiding effect of V 5 V. The network will
then appear as shown in Fig. 2.75 and vo vi 5 V. Substituting id 0 at vd 0 for
the transition levels, we obtain the network of Fig. 2.76 and vi 5 V.
Figure 2.75 vo with diode in
the “on” state.
– +
+
vd = 0 V
+
5 V id = 0 A
vi
vo = vR = iR R = id R = (0) R = 0 V
R
–
Figure 2.76 Determining the
transition level for the clipper of
Fig. 2.74.
–
For vi more negative than 5 V the diode will enter its open-circuit state, while
for voltages more positive than 5 V the diode is in the short-circuit state. The input
and output voltages appear in Fig. 2.77.
vo
vi
vi + 5 V = 20 V + 5 V = 25 V
20
vo = 0 V + 5 V = 5 V
5V
–5V
T
2
T
t
0
Transition
voltage
Figure 2.77
T
2
T
t
vo = –5 V + 5 V = 0 V
Sketching vo for Example 2.20.
The analysis of clipper networks with square-wave inputs is actually easier to analyze than with sinusoidal inputs because only two levels have to be considered. In
other words, the network can be analyzed as if it had two dc level inputs with the resulting output vo plotted in the proper time frame.
78
Chapter 2
Diode Applications
EXAMPLE 2.21
Repeat Example 2.20 for the square-wave input of Fig. 2.78.
Figure 2.78 Applied signal for
Example 2.21.
Solution
For vi 20 V (0 → T/ 2) the network of Fig. 2.79 will result. The diode is in the shortcircuit state and vo 20 V 5 V 25 V. For vi 10 V the network of Fig. 2.80
will result, placing the diode in the “off” state and vo iRR (0)R 0 V. The resulting output voltage appears in Fig. 2.81.
–
+
5V
+
+
R
20 V
–
–
vo
+
5V
vo = 0 V
R
10 V
+
–
Figure 2.79
– +
vo at vi 20 V.
–
Figure 2.80
vo at vi 10 V.
Figure 2.81 Sketching vo for
Example 2.21.
Note in Example 2.21 that the clipper not only clipped off 5 V from the total
swing but raised the dc level of the signal by 5 V.
Parallel
The network of Fig. 2.82 is the simplest of parallel diode configurations with the output for the same inputs of Fig. 2.67. The analysis of parallel configurations is very
similar to that applied to series configurations, as demonstrated in the next example.
+
vi
vi
vo
–
–
vo
vi
V
0
–V
Figure 2.82
+
R
vo
V
t
0
–V
t
0
t
0
t
–V
–V
Response to a parallel clipper.
2.9 Clippers
79
EXAMPLE 2.2
Determine vo for the network of Fig. 2.83.
Figure 2.83
Example 2.22.
Solution
The polarity of the dc supply and the direction of the diode strongly suggest that the
diode will be in the “on” state for the negative region of the input signal. For this region the network will appear as shown in Fig. 2.84, where the defined terminals for
vo require that vo V 4 V.
–
+
R
vi
vo = V = 4 V
V
+
4V
–
Figure 2.84
region of vi.
vo for the negative
The transition state can be determined from Fig. 2.85, where the condition id 0 A at vd 0 V has been imposed. The result is vi (transition) V 4 V.
Since the dc supply is obviously “pressuring” the diode to stay in the shortcircuit state, the input voltage must be greater than 4 V for the diode to be in the “off”
state. Any input voltage less than 4 V will result in a short-circuited diode.
For the open-circuit state the network will appear as shown in Fig. 2.86, where
vo vi. Completing the sketch of vo results in the waveform of Fig. 2.87.
Figure 2.85 Determining the
transition level for Example 2.22.
Figure 2.87 Sketching vo for
Example 2.22.
Figure 2.86 Determining vo for
the open state of the diode.
To examine the effects of VT on the output voltage, the next example will specify a silicon diode rather than an ideal diode equivalent.
80
Chapter 2
Diode Applications
Repeat Example 2.22 using a silicon diode with VT 0.7 V.
EXAMPLE 2.23
Solution
The transition voltage can first be determined by applying the condition id 0 A at
vd VD 0.7 V and obtaining the network of Fig. 2.88. Applying Kirchhoff’s voltage law around the output loop in the clockwise direction, we find that
vi VT V 0
and
vi V VT 4 V 0.7 V 3.3 V
Figure 2.88 Determining the
transition level for the network of
Fig. 2.83.
For input voltages greater than 3.3 V, the diode will be an open circuit and
vo vi. For input voltages of less than 3.3 V, the diode will be in the “on” state and
the network of Fig. 2.89 results, where
vo 4 V 0.7 V 3.3 V
Figure 2.89 Determining vo for
the diode of Fig. 2.83 in the “on”
state.
The resulting output waveform appears in Fig. 2.90. Note that the only effect of VT
was to drop the transition level to 3.3 from 4 V.
Figure 2.90 Sketching vo for
Example 2.23.
There is no question that including the effects of VT will complicate the analysis
somewhat, but once the analysis is understood with the ideal diode, the procedure,
including the effects of VT, will not be that difficult.
Summary
A variety of series and parallel clippers with the resulting output for the sinusoidal
input are provided in Fig. 2.91. In particular, note the response of the last configuration, with its ability to clip off a positive and a negative section as determined by the
magnitude of the dc supplies.
2.9 Clippers
81
Figure 2.91
82
Chapter 2
Clipping circuits.
Diode Applications
2.10 CLAMPERS
The clamping network is one that will “clamp” a signal to a different dc level. The
network must have a capacitor, a diode, and a resistive element, but it can also employ an independent dc supply to introduce an additional shift. The magnitude of R
and C must be chosen such that the time constant RC is large enough to ensure
that the voltage across the capacitor does not discharge significantly during the interval the diode is nonconducting. Throughout the analysis we will assume that for all
practical purposes the capacitor will fully charge or discharge in five time constants.
The network of Fig. 2.92 will clamp the input signal to the zero level (for ideal
diodes). The resistor R can be the load resistor or a parallel combination of the load
resistor and a resistor designed to provide the desired level of R.
C
+
V
+
V
–
+
R
vo
–
–
Figure 2.92
Clamper.
During the interval 0 → T/2 the network will appear as shown in Fig. 2.93, with
the diode in the “on” state effectively “shorting out” the effect of the resistor R. The
resulting RC time constant is so small (R determined by the inherent resistance of the
network) that the capacitor will charge to V volts very quickly. During this interval
the output voltage is directly across the short circuit and vo 0 V.
When the input switches to the V state, the network will appear as shown in
Fig. 2.94, with the open-circuit equivalent for the diode determined by the applied
signal and stored voltage across the capacitor—both “pressuring” current through the
diode from cathode to anode. Now that R is back in the network the time constant
determined by the RC product is sufficiently large to establish a discharge period 5
much greater than the period T/2 → T, and it can be assumed on an approximate basis that the capacitor holds onto all its charge and, therefore, voltage (since V Q/C)
during this period.
Since vo is in parallel with the diode and resistor, it can also be drawn in the alternative position shown in Fig. 2.94. Applying Kirchhoff’s voltage law around the
input loop will result in
Figure 2.93 Diode “on” and the
capacitor charging to V volts.
Figure 2.94 Determining vo
with the diode “off.”
V V vo 0
and
vo 2V
The negative sign resulting from the fact that the polarity of 2V is opposite to the polarity defined for vo. The resulting output waveform appears in Fig. 2.95 with the input signal. The output signal is clamped to 0 V for the interval 0 to T/2 but maintains
the same total swing (2V) as the input.
For a clamping network:
The total swing of the output is equal to the total swing of the input
signal.
This fact is an excellent checking tool for the result obtained.
In general, the following steps may be helpful when analyzing clamping networks:
1. Start the analysis of clamping networks by considering that part of the input signal that will forward bias the diode.
Figure 2.95 Sketching vo for the
network of Fig. 2.92.
2.10 Clampers
83
The statement above may require skipping an interval of the input signal (as demonstrated in an example to follow), but the analysis will not be extended by an unnecessary measure of investigation.
2. During the period that the diode is in the “on” state, assume that the capacitor will charge up instantaneously to a voltage level determined by the
network.
3. Assume that during the period when the diode is in the “off” state the capacitor will hold on to its established voltage level.
4. Throughout the analysis maintain a continual awareness of the location
and reference polarity for vo to ensure that the proper levels for vo are obtained.
5. Keep in mind the general rule that the total swing of the total output must
match the swing of the input signal.
EXAMPLE 2.24
Determine vo for the network of Fig. 2.96 for the input indicated.
Figure 2.96
Applied signal and network for Example 2.24.
Solution
Figure 2.97 Determining vo and
VC with the diode in the “on”
state.
Note that the frequency is 1000 Hz, resulting in a period of 1 ms and an interval of
0.5 ms between levels. The analysis will begin with the period t1 → t2 of the input
signal since the diode is in its short-circuit state as recommended by comment 1. For
this interval the network will appear as shown in Fig. 2.97. The output is across R,
but it is also directly across the 5-V battery if you follow the direct connection between the defined terminals for vo and the battery terminals. The result is vo 5 V
for this interval. Applying Kirchhoff’s voltage law around the input loop will result in
20 V VC 5 V 0
VC 25 V
and
The capacitor will therefore charge up to 25 V, as stated in comment 2. In this
case the resistor R is not shorted out by the diode but a Thévenin equivalent circuit
of that portion of the network which includes the battery and the resistor will result
in RTh 0 with ETh V 5 V. For the period t2 → t3 the network will appear as
shown in Fig. 2.98.
The open-circuit equivalent for the diode will remove the 5-V battery from having any effect on vo, and applying Kirchhoff’s voltage law around the outside loop of
the network will result in
10 V 25 V vo 0
Figure 2.98 Determining vo
with the diode in the “off” state.
and
84
Chapter 2
vo 35 V
Diode Applications
The time constant of the discharging network of Fig. 2.98 is determined by the
product RC and has the magnitude
RC (100 k)(0.1 F) 0.01 s 10 ms
The total discharge time is therefore 5 5(10 ms) 50 ms.
Since the interval t2 → t3 will only last for 0.5 ms, it is certainly a good approximation that the capacitor will hold its voltage during the discharge period between pulses
of the input signal. The resulting output appears in Fig. 2.99 with the input signal.
Note that the output swing of 30 V matches the input swing as noted in step 5.
Figure 2.99 vi and vo for the
clamper of Fig. 2.96.
Repeat Example 2.24 using a silicon diode with VT 0.7 V.
EXAMPLE 2.25
Solution
For the short-circuit state the network now takes on the appearance of Fig. 2.100 and
vo can be determined by Kirchhoff’s voltage law in the output section.
5 V 0.7 V vo 0
and
vo 5 V 0.7 V 4.3 V
For the input section Kirchhoff’s voltage law will result in
20 V VC 0.7 V 5 V 0
and
VC 25 V 0.7 V 24.3 V
For the period t2 → t3 the network will now appear as in Fig. 2.101, with the only
change being the voltage across the capacitor. Applying Kirchhoff’s voltage law yields
Figure 2.100 Determining vo
and VC with the diode in the “on”
state.
10 V 24.3 V vo 0
and
vo 34.3 V
Figure 2.101 Determining vo
with the diode in the open state.
2.10 Clampers
85
The resulting output appears in Fig. 2.102, verifying the statement that the input and
output swings are the same.
Figure 2.102 Sketching vo for
the clamper of Fig. 2.96 with a
silicon diode.
A number of clamping circuits and their effect on the input signal are shown in
Fig. 2.103. Although all the waveforms appearing in Fig. 2.103 are square waves,
clamping networks work equally well for sinusoidal signals. In fact, one approach to
the analysis of clamping networks with sinusoidal inputs is to replace the sinusoidal
signal by a square wave of the same peak values. The resulting output will then form
an envelope for the sinusoidal response as shown in Fig. 2.104 for a network appearing in the bottom right of Fig. 2.103.
Figure 2.103 Clamping circuits with ideal diodes (5 5RC T/2).
86
Chapter 2
Diode Applications
vo (V)
vi
+30
20 V
+
t
0
–20 V
+
C
–
vi
R
vo
10 V
+
–
t
0
–10 V
–
Figure 2.104 Clamping network with a sinusoidal input.
2.11 ZENER DIODES
The analysis of networks employing Zener diodes is quite similar to that applied to
the analysis of semiconductor diodes in previous sections. First the state of the diode
must be determined followed by a substitution of the appropriate model and a determination of the other unknown quantities of the network. Unless otherwise specified,
the Zener model to be employed for the “on” state will be as shown in Fig. 2.105a.
For the “off” state as defined by a voltage less than VZ but greater than 0 V with the
polarity indicated in Fig. 2.105b, the Zener equivalent is the open circuit that appears
in the same figure.
Figure 2.105 Zener diode
equivalents for the (a) “on” and
(b) “off” states.
V
i
and R
The simplest of Zener diode networks appears in Fig. 2.106. The applied dc voltage
is fixed, as is the load resistor. The analysis can fundamentally be broken down into
two steps.
1. Determine the state of the Zener diode by removing it from the network
and calculating the voltage across the resulting open circuit.
Figure 2.106 Basic Zener regulator.
Applying step 1 to the network of Fig. 2.106 will result in the network of Fig.
2.107, where an application of the voltage divider rule will result in
R Vi
V VL L
R RL
(2.16)
If V VZ, the Zener diode is “on” and the equivalent model of Fig. 2.105a can be
substituted. If V VZ, the diode is “off” and the open-circuit equivalence of Fig.
2.105b is substituted.
2.11 Zener Diodes
Figure 2.107 Determining the
state of the Zener diode.
87
2. Substitute the appropriate equivalent circuit and solve for the desired unknowns.
For the network of Fig. 2.106, the “on” state will result in the equivalent network
of Fig. 2.108. Since voltages across parallel elements must be the same, we find that
VL VZ
(2.17)
The Zener diode current must be determined by an application of Kirchhoff’s current
law. That is,
Figure 2.108 Substituting the
Zener equivalent for the “on” situation.
IR IZ IL
IZ IR IL
and
(2.18)
where
VL
IL RL
and
VR
Vi VL
IR R
R
The power dissipated by the Zener diode is determined by
PZ VZ IZ
(2.19)
which must be less than the PZM specified for the device.
Before continuing, it is particularly important to realize that the first step was employed only to determine the state of the Zener diode. If the Zener diode is in the
“on” state, the voltage across the diode is not V volts. When the system is turned on,
the Zener diode will turn “on” as soon as the voltage across the Zener diode is VZ
volts. It will then “lock in” at this level and never reach the higher level of V volts.
Zener diodes are most frequently used in regulator networks or as a reference
voltage. Figure 2.106 is a simple regulator designed to maintain a fixed voltage across
the load RL. For values of applied voltage greater than required to turn the Zener diode
“on,” the voltage across the load will be maintained at VZ volts. If the Zener diode is
employed as a reference voltage, it will provide a level for comparison against other
voltages.
EXAMPLE 2.26
(a) For the Zener diode network of Fig. 2.109, determine VL, VR, IZ, and PZ.
(b) Repeat part (a) with RL = 3 k.
Figure 2.109 Zener diode
regulator for Example 2.26.
Solution
(a) Following the suggested procedure the network is redrawn as shown in Fig.
2.110. Applying Eq. (2.16) gives
R Vi
1.2 k(16 V)
8.73 V
V L
R RL
1 k 1.2 k
88
Chapter 2
Diode Applications
Figure 2.110 Determining V for
the regulator of Fig. 2.109.
Since V 8.73 V is less than VZ 10 V, the diode is in the “off” state as shown
on the characteristics of Fig. 2.111. Substituting the open-circuit equivalent will result in the same network as in Fig. 2.110, where we find that
VL V 8.73 V
VR Vi VL 16 V 8.73 V 7.27 V
IZ 0 A
and
Figure 2.111 Resulting operating point for the network of Fig.
2.109.
PZ VZIZ VZ (0 A) 0 W
(b) Applying Eq. (2.16) will now result in
R Vi
3 k(16 V)
V L
12 V
R RL
1 k 3 k
Since V 12 V is greater than VZ 10 V, the diode is in the “on” state and the network of Fig. 2.112 will result. Applying Eq. (2.17) yields
VL VZ 10 V
VR Vi VL 16 V 10 V 6 V
and
with
VL
10 V
IL 3.33 mA
RL
3 k
and
VR
6V
IR 6 mA
R
1 k
so that
IZ IR IL [Eq. (2.18)]
6 mA 3.33 mA
2.67 mA
The power dissipated,
PZ VZIZ (10 V)(2.67 mA) 26.7 mW
which is less than the specified PZM 30 mW.
Figure 2.112 Network of Fig.
2.109 in the “on” state.
2.11 Zener Diodes
89
Fixed Vi, Variable RL
Due to the offset voltage VZ, there is a specific range of resistor values (and therefore
load current) which will ensure that the Zener is in the “on” state. Too small a load
resistance RL will result in a voltage VL across the load resistor less than VZ , and the
Zener device will be in the “off” state.
To determine the minimum load resistance of Fig. 2.106 that will turn the Zener
diode on, simply calculate the value of RL that will result in a load voltage VL VZ.
That is,
R Vi
VL VZ L
RL R
Solving for RL, we have
RVZ
RLmin Vi VZ
(2.20)
Any load resistance value greater than the RL obtained from Eq. (2.20) will ensure
that the Zener diode is in the “on” state and the diode can be replaced by its VZ source
equivalent.
The condition defined by Eq. (2.20) establishes the minimum RL but in turn specifies the maximum IL as
VL
VZ
ILmax RL
RLmin
(2.21)
Once the diode is in the “on” state, the voltage across R remains fixed at
VR Vi VZ
(2.22)
VR
IR R
(2.23)
IZ IR IL
(2.24)
and IR remains fixed at
The Zener current
resulting in a minimum IZ when IL is a maximum and a maximum IZ when IL is a
minimum value since IR is constant.
Since IZ is limited to IZM as provided on the data sheet, it does affect the range
of RL and therefore IL. Substituting IZM for IZ establishes the minimum IL as
ILmin IR IZM
(2.25)
and the maximum load resistance as
VZ
RLmax ILmin
90
Chapter 2
Diode Applications
(2.26)
(a) For the network of Fig. 2.113, determine the range of RL and IL that will result
in VRL being maintained at 10 V.
(b) Determine the maximum wattage rating of the diode.
EXAMPLE 2.27
Figure 2.113 Voltage regulator
for Example 2.27.
Solution
(a) To determine the value of RL that will turn the Zener diode on, apply Eq. (2.20):
RVZ
(1 k)(10 V)
10 k
RLmin 250 Vi VZ
50 V 10 V
40
The voltage across the resistor R is then determined by Eq. (2.22):
VR Vi VZ 50 V 10 V 40 V
and Eq. (2.23) provides the magnitude of IR:
VR
40 V
IR 40 mA
R
1 k
The minimum level of IL is then determined by Eq. (2.25):
ILmin IR IZM 40 mA 32 mA 8 mA
with Eq. (2.26) determining the maximum value of RL:
VZ
10 V
RLmax 1.25 k
ILmin
8 mA
A plot of VL versus RL appears in Fig. 2.114a and for VL versus IL in Fig. 2.114b.
(b) Pmax VZ IZM
(10 V)(32 mA) 320 mW
Figure 2.114 VL versus RL and IL for the regulator of Fig. 2.113.
2.11 Zener Diodes
91
Fixed RL, Variable Vi
For fixed values of RL in Fig. 2.106, the voltage Vi must be sufficiently large to turn
the Zener diode on. The minimum turn-on voltage Vi Vimin is determined by
R Vi
VL VZ L
RL R
(RL R)VZ
Vimin RL
and
(2.27)
The maximum value of Vi is limited by the maximum Zener current IZM. Since
IZM IR IL,
IRmax IZM IL
(2.28)
Since IL is fixed at VZ /RL and IZM is the maximum value of IZ, the maximum Vi
is defined by
Vimax VRmax VZ
Vimax IRmaxR VZ
EXAMPLE 2.28
(2.29)
Determine the range of values of Vi that will maintain the Zener diode of Fig. 2.115
in the “on” state.
Figure 2.115 Regulator for Example 2.28.
Solution
(RL R)VZ (1200 220 )(20 V)
Eq. (2.27): Vimin 23.67 V
1200 RL
VL
VZ
20 V
16.67 mA
IL 1.2 k
RL RL
Eq. (2.28): IRmax IZM IL 60 mA 16.67 mA
76.67 mA
Eq. (2.29): Vimax IRmaxR VZ
(76.67 mA)(0.22 k) 20 V
16.87 V 20 V
36.87 V
Figure 2.116 VL versus Vi for
the regulator of Fig. 2.115.
92
A plot of VL versus Vi is provided in Fig. 2.116.
Chapter 2
Diode Applications
The results of Example 2.28 reveal that for the network of Fig. 2.115 with a fixed
RL, the output voltage will remain fixed at 20 V for a range of input voltage that extends from 23.67 to 36.87 V.
In fact, the input could appear as shown in Fig. 2.117 and the output would remain constant at 20 V, as shown in Fig. 2.116. The waveform appearing in Fig. 2.117
is obtained by filtering a half-wave- or full-wave-rectified output—a process described
in detail in a later chapter. The net effect, however, is to establish a steady dc voltage
(for a defined range of Vi) such as that shown in Fig. 2.116 from a sinusoidal source
with 0 average value.
Figure 2.117 Waveform generated by a filtered rectified signal.
Two or more reference levels can be established by placing Zener diodes in series
as shown in Fig. 2.118. As long as Vi is greater than the sum of VZ1 and VZ2, both
diodes will be in the “on” state and the three reference voltages will be available.
Two back-to-back Zeners can also be used as an ac regulator as shown in Fig.
2.119a. For the sinusoidal signal vi the circuit will appear as shown in Fig. 2.119b at
the instant vi 10 V. The region of operation for each diode is indicated in the adjoining figure. Note that Z1 is in a low-impedance region, while the impedance of Z2
is quite large, corresponding with the open-circuit representation. The result is that
vo vi when vi 10 V. The input and output will continue to duplicate each other
until vi reaches 20 V. Z2 will then “turn on” (as a Zener diode), while Z1 will be in a
vi
+
vo
+
5 kΩ
22 V
Figure 2.118 Establishing three
reference voltage levels.
Z1
ωt
0
–22 V
vi
vo
20-V
Zeners
Z2
–
20 V
0
20 V
ωt
–
(a)
I
+
5 kΩ
–
+
vi = 10 V
–
Z1
20 V
0
V
Z2
(b)
Figure 2.119 Sinusoidal ac regulation: (a) 40-V peak-to-peak sinusoidal ac regulator; (b) circuit operation at vi 10 V.
2.11 Zener Diodes
93
region of conduction with a resistance level sufficiently small compared to the series
5-k resistor to be considered a short circuit. The resulting output for the full range
of vi is provided in Fig. 2.119(a). Note that the waveform is not purely sinusoidal, but
its rms value is lower than that associated with a full 22-V peak signal. The network
is effectively limiting the rms value of the available voltage. The network of Fig.
2.119a can be extended to that of a simple square-wave generator (due to the clipping action) if the signal vi is increased to perhaps a 50-V peak with 10-V Zeners as
shown in Fig. 2.120 with the resulting output waveform.
vi
50 V
0
+
5 kΩ
vi
10-V
Zeners
vo
+
+
Z1
π
2π ω t
–
+
vo
–10 V
Z2
–
–
10 V
–
Figure 2.120 Simple square-wave generator.
2.12 VOLTAGE-MULTIPLIER CIRCUITS
Voltage-multiplier circuits are employed to maintain a relatively low transformer peak
voltage while stepping up the peak output voltage to two, three, four, or more times
the peak rectified voltage.
Voltage Doubler
The network of Figure 2.121 is a half-wave voltage doubler. During the positive voltage half-cycle across the transformer, secondary diode D1 conducts (and diode D2 is
cut off), charging capacitor C1 up to the peak rectified voltage (Vm). Diode D1 is ideally a short during this half-cycle, and the input voltage charges capacitor C1 to Vm
with the polarity shown in Fig. 2.122a. During the negative half-cycle of the secondary voltage, diode D1 is cut off and diode D2 conducts charging capacitor C2.
Since diode D2 acts as a short during the negative half-cycle (and diode D1 is open),
we can sum the voltages around the outside loop (see Fig. 2.122b):
Vm VC1 VC2 0
Vm Vm VC2 0
from which
VC 2 2Vm
Figure 2.121 Half-wave voltage
doubler.
94
Chapter 2
Diode Applications
Figure 2.122 Double operation, showing each half-cycle of
operation: (a) positive half-cycle;
(b) negative half cycle.
On the next positive half-cycle, diode D2 is nonconducting and capacitor C2 will discharge through the load. If no load is connected across capacitor C2, both capacitors
stay charged—C1 to Vm and C2 to 2Vm. If, as would be expected, there is a load connected to the output of the voltage doubler, the voltage across capacitor C2 drops during the positive half-cycle (at the input) and the capacitor is recharged up to 2Vm during the negative half-cycle. The output waveform across capacitor C2 is that of a
half-wave signal filtered by a capacitor filter. The peak inverse voltage across each
diode is 2Vm.
Another doubler circuit is the full-wave doubler of Fig. 2.123. During the positive half-cycle of transformer secondary voltage (see Fig. 2.124a) diode D1 conducts
charging capacitor C1 to a peak voltage Vm. Diode D2 is nonconducting at this time.
Figure 2.123 Full-wave voltage
doubler.
D1
Conducting
+
Nonconducting
–
C1
Vm
D1
+
–
Vm
–
C1
Vm
+
–
Vm
+
+
C2
D2
(a)
–
+
Vm
C2
D2
Nonconducting
–
Vm
Conducting
(b)
2.12 Voltage-Multiplier Circuits
Figure 2.124 Alternate halfcycles of operation for full-wave
voltage doubler.
95
During the negative half-cycle (see Fig. 2.124b) diode D2 conducts charging capacitor C2 while diode D1 is nonconducting. If no load current is drawn from the circuit, the voltage across capacitors C1 and C2 is 2Vm. If load current is drawn from the
circuit, the voltage across capacitors C1 and C2 is the same as that across a capacitor
fed by a full-wave rectifier circuit. One difference is that the effective capacitance
is that of C1 and C2 in series, which is less than the capacitance of either C1 or C2
alone. The lower capacitor value will provide poorer filtering action than the singlecapacitor filter circuit.
The peak inverse voltage across each diode is 2Vm , as it is for the filter capacitor
circuit. In summary, the half-wave or full-wave voltage-doubler circuits provide twice
the peak voltage of the transformer secondary while requiring no center-tapped transformer and only 2Vm PIV rating for the diodes.
Voltage Tripler and Quadrupler
Figure 2.125 shows an extension of the half-wave voltage doubler, which develops
three and four times the peak input voltage. It should be obvious from the pattern of
the circuit connection how additional diodes and capacitors may be connected so that
the output voltage may also be five, six, seven, and so on, times the basic peak
voltage (Vm).
Figure 2.125 Voltage tripler and quadrupler.
In operation capacitor C1 charges through diode D1 to a peak voltage, Vm, during
the positive half-cycle of the transformer secondary voltage. Capacitor C2 charges to
twice the peak voltage 2Vm developed by the sum of the voltages across capacitor C1
and the transformer, during the negative half-cycle of the transformer secondary voltage.
During the positive half-cycle, diode D3 conducts and the voltage across capacitor C2 charges capacitor C3 to the same 2Vm peak voltage. On the negative halfcycle, diodes D2 and D4 conduct with capacitor C3, charging C4 to 2Vm.
The voltage across capacitor C2 is 2Vm, across C1 and C3 it is 3Vm, and across C2
and C4 it is 4Vm. If additional sections of diode and capacitor are used, each capacitor will be charged to 2Vm. Measuring from the top of the transformer winding (Fig.
2.125) will provide odd multiples of Vm at the output, whereas measuring the output
voltage from the bottom of the transformer will provide even multiples of the peak
voltage, Vm.
The transformer rating is only Vm, maximum, and each diode in the circuit must
be rated at 2Vm PIV. If the load is small and the capacitors have little leakage, extremely high dc voltages may be developed by this type of circuit, using many sections to step up the dc voltage.
96
Chapter 2
Diode Applications
2.13 PSPICE WINDOWS
Series Diode Configuration
PSpice Windows will now be applied to the network of Fig. 2.27 to permit a comparison with the hand-calculated solution. As briefly described in Chapter 1, the application of PSpice Windows requires that the network first be constructed on the
schematics screen. The next few paragraphs will examine the basics of setting up the
network on the screen, assuming no prior experience with the process. It might be
helpful to reference the completed network of Fig. 2.126 as we progress through the
discussion.
Figure 2.126 PSpice Windows
analysis of a series diode
configuration.
In general, it is easier to draw the network if the grid is on the screen and the stipulation is made that all elements be on the grid. This will ensure that all the connections are made between the elements. The screen can be set up by first choosing Options at the heading of the schematics screen, followed by Display Options. The
Display Options dialog box will permit you to make all the choices necessary regarding the type of display desired. For our purposes, we will choose Grid On, Stay
on Grid, and Grid Spacing of 0.1 in.
R
The resistor R will be the first to be positioned. By clicking on the Get New Part
icon (the icon in the top right area with the binoculars) followed by Libraries, we
can choose the Analog.slb library of basic elements. We can then scroll the Part list
until we find R. Clicking on R followed by OK will result in the Part Browser Basic dialog box reflecting our choice of a resistive element. Choosing the Place &
Close option will place the resistive element on the screen and close the dialog box.
The resistor will appear horizontal, which is perfect for the R1 of Fig. 2.27 (note Fig.
2.126). Move the resistor to a logical location, and click the left button of the mouse—
the resistor R1 is in place. Note that it snaps to the grid structure. The resistor R2 must
now be placed to the right of R1. By simply moving the mouse to the right, the second resistor will appear, and R2 can be placed in the proper location with a subsequent click of the mouse. Since the network only has two resistors, the depositing of
resistors can be ended by a right click of the mouse. The resistor R2 can be rotated
by pressing the keys Ctrl and R simultaneously or by choosing Edit on the menu
bar, followed by Rotate.
The result of the above is two resistors with the right labels but the wrong values. To change a value, double click on the value of the screen (first R1). A Set Attribute Value dialog box will appear. Type in the correct value, and send the value
to the screen with OK. The 4.7k will appear within a box that can be moved by
simply clicking on the small box and, while holding the clicker down, moving the
4.7k to the desired location. Release the clicker, and the 4.7k label will remain
where placed. Once located, an additional click anywhere on the screen will remove
the boxes and end the process. If you want to move the 4.7k in the future, simply
click once on the value and the boxes will reappear. Repeat the above for the value
of the resistor R2.
2.13 PSpice Windows
97
To remove (clip) an element, simply click on it (to establish the red or active
color), and then click the scissors icon or use the sequence Edit-Delete.
E
The voltage sources are set by going to the source.slb library of Library Browser
and choosing VDC. Clicking OK results in the source symbol appearing on the
schematic. This symbol can be placed as required. After clicking it in the appropriate place, a V1 label will appear. To change the label to E1 simply click the V1 twice
and an Edit Reference Designator dialog box will appear. Change the label to E1
and click OK, and then E1 will appear on the screen within a box. The box can be
moved in the same manner as the labels for resistors. When you have the correct position, simply click the mouse once more and place E1 as desired.
To set the value of E1, click the value twice and the Set Attribute Value will appear. Set the value to 10V and click OK. The new value will appear on the schematic.
The value can also be set by clicking the battery symbol itself twice, after which a
dialog box will appear labeled E1 PartName:VDC. By choosing DC 0V, DC and
Value will appear in the designated areas at the top of the dialog box. Using the
mouse, bring the marker to the Value box and change it to 10V. Then click Save Attr.
to be sure and save the new value, and an OK will result in E1 being changed to 10V.
E1 can now be set, but be sure to turn it 180° with the appropriate operations.
DIODE
The diode is found in the EVAL.slb library of the Library Browser dialog box.
Choosing the D1N4148 diode followed by an OK and Close & Place will place the
diode symbol on the screen. Move the diode to the correct position, click it in place
with a left click, and end the operation with a right click of the mouse. The labels D1
and D1N4148 will appear near the diode. Clicking on either label will provide the
boxes that permit movement of the labels.
Let us now take a look at the diode specs by clicking the diode symbol once, followed by the Edit-Model-Edit Instance Model sequence. For the moment, we will
leave the parameters as listed. In particular, note that Is 2.682nA and the terminal
capacitance (important when the applied frequency becomes a factor) is 4pF.
IPROBE
One or more currents of a network can be displayed by inserting an IPROBE in
the desired path. IPROBE is found in the SPECIAL.slb library and appears as a meter face on the screen. IPROBE will respond with a positive answer if the current
(conventional) enters the symbol at the end with the arc representing the scale. Since
we are looking for a positive answer in this investigation, IPROBE should be installed as shown in Fig. 2.126. When the symbol first appears, it is 180° out of phase
with the desired current. Therefore, it is necessary to use the Ctrl-R sequence twice
to rotate the symbol before finalizing its position. As with the elements described
above, once it is in place a single click will place the meter and a right click will complete the insertion process.
LINE
The elements now need to be connected by choosing the icon with the thin line
and pencil or by the sequence Draw-Wire. A pencil will appear that can draw the desired connections in the following manner: Move the pencil to the beginning of the
line, and click the left side of the mouse. The pencil is now ready to draw. Draw the
desired line (connection), and click the left side again when the connection is complete. The line will appear in red, waiting for another random click of the mouse or
98
Chapter 2
Diode Applications
the insertion of another line. It will then turn geen to indicate it is in memory. For
additional lines, simply repeat the procedure. When done, simply click the right side
of the mouse.
EGND
The system must have a ground to serve as a reference point for the nodal voltages. Earth ground (EGND) is part of the PORT.slb library and can be placed in the
same manner as the elements described above.
VIEWPOINT
Nodal voltages can be displayed on the diagram after the simulation using VIEWPOINTS, which is found in the SPECIAL.slb library. Simply place the arrow of the
VIEWPOINT symbol where you desire the voltage with respect to ground. A VIEWPOINT can be placed at every node of the network if necessary, although only three
are placed in Fig. 2.126. The network is now complete, as shown in Fig 2.126.
ANALYSIS
The network is now ready to be analyzed. To expedite the process, click on Analysis and choose Probe Setup. By selecting Do not auto-run Probe you save intermediary steps that are inappropriate for this analysis; it is an option that will be discussed later in this chapter. After OK, go to Analysis and choose Simulation. If the
network was installed properly, a PSpiceAD dialog box will appear and reveal that
the bias (dc) points have been calculated. If we now exit the box by clicking on the
small x in the top right corner, you will obtain the results appearing in Fig. 2.126.
Note that the program has automatically provided four dc voltages of the network
(in addition to the VIEWPOINT voltages). This occurred because an option under
analysis was enabled. For future analysis we will want control over what is displayed
so follow the path through Analysis-Display Results on Schematic and slide over to
the adjoining Enable box. Clicking the Enable box will remove the check, and the
dc voltages will not automatically appear. They will only appear where VIEWPOINTS have been inserted. A more direct path toward controlling the appearance
of the dc voltages is to use the icon on the menu bar with the large capital V. By clicking it on and off, you can control whether the dc levels of the network will appear.
The icon with the large capital I will permit all the dc currents of the network to be
shown if desired. For practice, click it on and off and note the effect on the schematic.
If you want to remove selected dc voltages on the schematic, simply click the nodal
voltage of interest, then click the icon with the smaller capital V in the same grouping. Clicking it once will remove the selected dc voltage. The same can be done for
selected currents with the remaining icon of the group. For the future, it should be
noted that an analysis can also be initiated by simply clicking the Simulation icon
having the yellow background and the two waveforms (square wave and sinusoidal).
Note also that the results are not an exact match with those obtained in Example
2.11. The VIEWPOINT voltage at the far right is 421.56 rather than the 454.2
mV obtained in Example 2.11. In addition, the current is 2.081 rather than the 2.066
mA obtained in the same example. Further, the voltage across the diode is 281.79 mV
421.56 mV 0.64 V rather than the 0.7 V assumed for all silicon diodes. This all
results from our using a real diode with a long list of variables defining its operation.
However, it is important to remember that the analysis of Example 2.11 was an approximate one and, therefore, it is expected that the results are only close to the actual response. On the other hand, the results obtained for the nodal voltage and current are quite close. If taken to the tenths place, the currents (2.1 mA) are an exact match.
The results obtained in Fig. 2.126 can be improved (in the sense that they will be
a closer match to the hand-written solution) by clicking on the diode (to make it red)
2.13 PSpice Windows
99
Figure 2.127 The circuit of Figure 2.126 reexamined with
Is set to 3.5E-15A.
and using the sequence Edit-Model-Edit Instance Model (Text) to obtain the Model
Editor dialog box. Choose Is 3.5E-15A (a value determined by trial and error),
and delete all the other parameters for the device. Then, follow with OK-Simulate
icon to obtain the results of Fig. 2.127. Note that the voltage across the diode now is
260.17 mV 440.93 mV 0.701 V, or almost exactly 0.7 V. The VIEWPOINT voltage is 440.93 V or, again, an almost perfect match with the hand-written solution
of 0.44 V. In either case, the results obtained are very close to the expected values.
One is more accurate as far as the actual device is concerned, while the other provides
an almost exact match with the hand-written solution. One cannot expect a perfect
match for every diode network by simply setting Is to 3.5E-15A. As the current through
the diode changes, the level of Is must also change if an exact match with the handwritten solution is to be obtained. However, rather than worry about the current in
each system, it is suggested that Is 3.5E-15A be used as the standard value if the
PSpice solution is desired to be a close match with the hand-written solution. The results will not always be perfect, but in most cases they will be closer than if the parameters of the diode are left at their default values. For transistors in the chapters to
follow, it will be set to 2E-15A to obtain a suitable match with the hand-written solution. Note also that the Bias Current Display was enabled to show that the current
is indeed the same everywhere in the circuit.
The results can also be viewed in tabulated form by returning to Analysis and
choosing Examine Output. The result is the long listing of Fig. 2.128. The Schematics Netlist describes the network in terms of numbered nodes. The 0 refers to ground
level, with the 10V source from node 0 to 5. The source E2 is from 0 to node 3. The
resistor R2 is connected from node 3 to 4, and so on. Scrolling down the output file,
we find the Diode MODEL PARAMETERS clearly showing that Is is set at 3.5E15A and is the only parameter listed. Next is the SMALL SIGNAL BIAS SOLUTION or dc solution with the voltages at the various nodes. In addition, the current
through the sources of the network is shown. The negative sign reveals that it is reflecting the direction of electron flow (into the positive terminal). The total power dissipation of the elements is 31.1 mW. Finally, the OPERATING POINT INFORMATION reveals that the current through the diode is 2.07 mA and the voltage across
the diode 0.701 V.
The analysis is now complete for the diode circuit of interest. We have not touched
on all the alternative paths available through PSpice Windows, but sufficient coverage has been provided to examine any of the networks covered in this chapter with a
dc source. For practice, the other examples should be examined using the Windows
approach since the results are provided for comparison. The same can be said for the
odd-numbered exercises at the end of this chapter.
Diode Characteristics
The characteristics of the D1N4148 diode used in the above analysis will now be obtained using a few maneuvers somewhat more sophisticated than those employed previously. First, the network in Fig. 2.129 is constructed using the procedures described
100
Chapter 2
Diode Applications
Figure 2.128 Output file for
PSpice Windows analysis of the
circuit of Figure 2.127.
above. Note, however, the Vd appearing above the diode D1. A point in the network
(representing the voltage from anode to ground for the diode) has been identified as
a particular voltage by double-clicking on the wire above the device and typing Vd
in the Set Attribute Value as the LABEL. The resulting voltage Vd is, in this case,
the voltage across the diode.
Next, Analysis Setup is chosen by either clicking on the Analysis Setup icon (at
the top left edge of the schematic with the horizontal blue bar and the two small
squares and rectangles) or by using the sequence Analysis-Setup. Within the Analysis-Setup dialog box the DC Sweep is enabled (the only one necessary for this exercise), followed by a single click of the DC Sweep rectangle. The DC Sweep dialog box will appear with various inquiries. In this case, we plan to sweep the source
voltage from 0 to 10 V in 0.01-V increments, so the Swept Var. Type is Voltage
Source, the Sweep Type will be linear, the Name E, and the Start Value 0V, the End
Value 10V, and the Increment 0.01V. Then, with an OK followed by a Close of the
2.13 PSpice Windows
Figure 2.129 Network to obtain the characteristics of the
D1N4148 diode.
101
Analysis Setup box, we are set to obtain the solution. The analysis to be performed
will obtain a complete solution for the network for each value of E from 0 to 10 V
in 0.01-V increments. In other words, the network will be analyzed 1000 times and
the resulting data stored for the plot to be obtained. The analysis is performed by the
sequence Analysis-Run Probe, followed by an immediate appearance of the MicroSim Probe graph showing only a horizontal axis of the source voltage E running
from 0 to 10 V.
Figure 2.130 Characteristics of
the D1N4148 diode.
Since the plot we want is of ID versus VD, we have to change the horizontal
(x-axis) to VD. This is accomplished by selecting Plot and then X-Axis Settings to
obtain the X Axis Settings dialog box. Next, we click Axis Variable and select V(Vd)
from the listing. After OK, we return to the dialog box to set the horizontal scale.
Choose User Defined, then enter 0V to 1V since this is the range of interest for Vd
with a Linear scale. Click OK and you will find that the horizontal axis is now V(Vd)
with a range of 0 to 1.0 V. The vertical axis must now be set to ID by first choosing
Trace (or the Trace icon, which is the red waveform with two sharp peaks and a set
of axis) and then Add to obtain Add Traces. Choosing I(D1) and clicking OK will
result in the plot of Fig. 2.130. In this case, the resulting plot extended from 0 to 10 mA.
The range can be reduced or expanded by simply going to Plot-Y-Axis Setting and
defining the range of interest.
In the previous analysis, the voltage across the diode was 0.64 V, corresponding
to a current of about 2 mA on the graph (recall the solution of 2.07 mA for the current). If the resulting current had been closer to 6.5 mA, the voltage across the diode
would have been about 0.7 V and the PSpice solution closer to the hand-written approach. If Is had been set to 3.5E-15A and all other parameters removed from the
diode listing, the curve would have shifted to the right and an intersection of 0.7 V
and 2.07 mA would have obtained.
102
Chapter 2
Diode Applications
PROBLEMS
§ 2.2 Load-Line Analysis
1. (a) Using the characteristics of Fig. 2.131b, determine ID, VD, and VR for the circuit of Fig.
2.131a.
(b) Repeat part (a) using the approximate model for the diode and compare results.
(c) Repeat part (a) using the ideal model for the diode and compare results.
Figure 2.131 Problems 1, 2
2. (a)
(b)
(c)
(d)
Using the characteristics of Fig. 2.131b, determine ID and VD for the circuit of Fig. 2.132.
Repeat part (a) with R 0.47 k.
Repeat part (a) with R 0.18 k.
Is the level of VD relatively close to 0.7 V in each case?
Figure 2.132 Problems 2, 3
How do the resulting levels of ID compare? Comment accordingly.
3. Determine the value of R for the circuit of Fig. 2.132 that will result in a diode current of 10
mA if E 7 V. Use the characteristics of Fig. 2.131b for the diode.
4. (a) Using the approximate characteristics for the Si diode, determine the level of VD, ID, and
VR for the circuit of Fig. 2.133.
(b) Perform the same analysis as part (a) using the ideal model for the diode.
(c) Do the results obtained in parts (a) and (b) suggest that the ideal model can provide a good
approximation for the actual response under some conditions?
Problems
Figure 2.133 Problem 4
103
§ 2.4 Series Diode Configurations with DC Inputs
5. Determine the current I for each of the configurations of Fig. 2.134 using the approximate equivalent model for the diode.
Figure 2.134 Problem 5
6. Determine Vo and ID for the networks of Fig. 2.135.
Figure 2.135 Problems 6, 49
* 7. Determine the level of Vo for each network of Fig. 2.136.
Figure 2.136 Problem 7
* 8. Determine Vo and ID for the networks of Fig. 2.137.
Figure 2.137 Problem 8
104
Chapter 2
Diode Applications
* 9. Determine Vo1 and Vo2 for the networks of Fig. 2.138.
Figure 2.138 Problem 9
§ 2.5 Parallel and Series–Parallel Configurations
10. Determine Vo and ID for the networks of Fig. 2.139.
Figure 2.139 Problems 10, 50
* 11. Determine Vo and I for the networks of Fig. 2.140.
Figure 2.140 Problem 11
Problems
105
12. Determine Vo1, Vo2, and I for the network of Fig. 2.141.
* 13. Determine Vo and ID for the network of Fig. 2.142.
Figure 2.141 Problem 12
Figure 2.142 Problems 13, 51
§ 2.6 AND/OR Gates
14. Determine Vo for the network of Fig. 2.38 with 0 V on both inputs.
15. Determine Vo for the network of Fig. 2.38 with 10 V on both inputs.
16. Determine Vo for the network of Fig. 2.41 with 0 V on both inputs.
17. Determine Vo for the network of Fig. 2.41 with 10 V on both inputs.
18. Determine Vo for the negative logic OR gate of Fig. 2.143.
19. Determine Vo for the negative logic AND gate of Fig. 2.144.
20. Determine the level of Vo for the gate of Fig. 2.145.
Figure 2.143 Problem 18
21. Determine Vo for the configuration of Fig. 2.146.
Figure 2.144 Problem 19
Figure 2.145 Problem 20
Figure 2.146
Problem 21
§ 2.7 Sinusoidal Inputs; Half-Wave Rectification
22. Assuming an ideal diode, sketch vi, vd, and id for the half-wave rectifier of Fig. 2.147. The input is a sinusoidal waveform with a frequency of 60 Hz
* 23. Repeat Problem 22 with a silicon diode (VT 0.7 V).
* 24. Repeat Problem 22 with a 6.8-k load applied as shown in Fig. 2.148. Sketch vL and iL.
25. For the network of Fig. 2.149, sketch vo and determine Vdc.
Figure 2.147 Problems 22, 23,
24
Figure 2.148 Problem 24
106
Chapter 2
Diode Applications
Figure 2.149 Problem 25
* 26. For the network of Fig. 2.150, sketch vo and iR.
Figure 2.150 Problem 26
* 27. (a) Given Pmax 14 mW for each diode of Fig. 2.151, determine the maximum current rating
of each diode (using the approximate equivalent model).
(b) Determine Imax for Vimax 160 V.
(c) Determine the current through each diode at Vimax using the results of part (b).
(e) If only one diode were present, determine the diode current and compare it to the maximum
rating.
Figure 2.151 Problem 27
§ 2.8 Full-Wave Rectification
28. A full-wave bridge rectifier with a 120-V rms sinusoidal input has a load resistor of 1 k.
(a) If silicon diodes are employed, what is the dc voltage available at the load?
(b) Determine the required PIV rating of each diode.
(c) Find the maximum current through each diode during conduction.
(d) What is the required power rating of each diode?
29. Determine vo and the required PIV rating of each diode for the configuration of Fig. 2.152.
Figure 2.152 Problem 29
Problems
107
* 30. Sketch vo for the network of Fig. 2.153 and determine the dc voltage available.
Figure 2.153 Problem 30
* 31. Sketch vo for the network of Fig. 2.154 and determine the dc voltage available.
Figure 2.154 Problem 31
§ 2.9 Clippers
32. Determine vo for each network of Fig. 2.155 for the input shown.
Figure 2.155 Problem 32
33. Determine vo for each network of Fig. 2.156 for the input shown.
Figure 2.156 Problem 33
108
Chapter 2
Diode Applications
* 34. Determine vo for each network of Fig. 2.157 for the input shown.
Figure 2.157 Problem 34
* 35. Determine vo for each network of Fig. 2.158 for the input shown.
Figure 2.158 Problem 35
36. Sketch iR and vo for the network of Fig. 2.159 for the input shown.
Figure 2.159 Problem 36
§ 2.10 Clampers
37. Sketch vo for each network of Fig. 2.160 for the input shown.
Figure 2.160 Problem 37
Problems
109
38. Sketch vo for each network of Fig. 2.161 for the input shown. Would it be a good approximation to consider the diode to be ideal for both configurations? Why?
Figure 2.161 Problem 38
* 39. For the network of Fig. 2.162:
(a)
Calculate 5 .
(b) Compare 5 to half the period of the applied signal.
(c)
Sketch vo.
Figure 2.162 Problem 39
* 40. Design a clamper to perform the function indicated in Fig. 2.163.
Figure 2.163 Problem 40
* 41. Design a clamper to perform the function indicated in Fig. 2.164.
Figure 2.164 Problem 41
110
Chapter 2
Diode Applications
§ 2.11 Zener Diodes
* 42. (a) Determine VL, IL, IZ, and IR for the network Fig. 2.165 if RL 180 (b) Repeat part (a) if RL 470 .
(c) Determine the value of RL that will establish maximum power conditions for the Zener
diode.
(d) Determine the minimum value of RL to ensure that the Zener diode is in the “on” state.
Figure 2.165 Problem 42
* 43. (a) Design the network of Fig. 2.166 to maintain VL at 12 V for a load variation (IL) from 0
to 200 mA. That is, determine Rs and VZ.
(b) Determine PZmax for the Zener diode of part (a).
* 44. For the network of Fig. 2.167, determine the range of Vi that will maintain VL at 8 V and not
exceed the maximum power rating of the Zener diode.
45. Design a voltage regulator that will maintain an output voltage of 20 V across a 1-k load with
an input that will vary between 30 and 50 V. That is, determine the proper value of Rs and the
maximum current IZM.
Figure 2.166 Problem 43
46. Sketch the output of the network of Fig. 2.120 if the input is a 50-V square wave. Repeat for
a 5-V square wave.
§ 2.12 Voltage-Multiplier Circuits
47. Determine the voltage available from the voltage doubler of Fig. 2.121 if the secondary voltage of the transformer is 120 V (rms).
48. Determine the required PIV ratings of the diodes of Fig. 2.121 in terms of the peak secondary
voltage Vm.
§ 2.13 PSpice Windows
Figure 2.167 Problems 44, 52
49. Perform an analysis of the network of Fig. 2.135 using PSpice Windows.
50. Perform an analysis of the network of Fig. 2.139 using PSpice Windows.
51. Perform an analysis of the network of Fig. 2.142 using PSpice Windows.
52. Perform a general analysis of the Zener network of Fig. 2.167 using PSpice Windows.
*
Please Note: Asterisks indicate more difficult problems.
Problems
111
CHAPTER
3
Bipolar Junction
Transistors
3.1 INTRODUCTION
During the period 1904–1947, the vacuum tube was undoubtedly the electronic device of interest and development. In 1904, the vacuum-tube diode was introduced by
J. A. Fleming. Shortly thereafter, in 1906, Lee De Forest added a third element, called
the control grid, to the vacuum diode, resulting in the first amplifier, the triode. In
the following years, radio and television provided great stimulation to the tube industry. Production rose from about 1 million tubes in 1922 to about 100 million in
1937. In the early 1930s the four-element tetrode and five-element pentode gained
prominence in the electron-tube industry. In the years to follow, the industry became
one of primary importance and rapid advances were made in design, manufacturing
techniques, high-power and high-frequency applications, and miniaturization.
On December 23, 1947, however, the electronics industry was to experience the
advent of a completely new direction of interest and development. It was on the afternoon of this day that Walter H. Brattain and John Bardeen demonstrated the amplifying action of the first transistor at the Bell Telephone Laboratories. The original
transistor (a point-contact transistor) is shown in Fig. 3.1. The advantages of this threeterminal solid-state device over the tube were immediately obvious: It was smaller
Co-inventors of the first transistor
at Bell Laboratories: Dr. William
Shockley (seated); Dr. John
Bardeen (left); Dr. Walter H. Brattain. (Courtesy of AT&T
Archives.)
Dr. Shockley Born: London,
England, 1910
PhD Harvard,
1936
Dr. Bardeen Born: Madison,
Wisconsin, 1908
PhD Princeton,
1936
Dr. Brattain Born: Amoy, China,
1902
PhD University of
Minnesota, 1928
All shared the Nobel Prize in
1956 for this contribution.
112
Figure 3.1
The first transistor. (Courtesy Bell Telephone Laboratories.)
and lightweight; had no heater requirement or heater loss; had rugged construction;
and was more efficient since less power was absorbed by the device itself; it was instantly available for use, requiring no warm-up period; and lower operating voltages
were possible. Note in the discussion above that this chapter is our first discussion of
devices with three or more terminals. You will find that all amplifiers (devices that
increase the voltage, current, or power level) will have at least three terminals with
one controlling the flow between two other terminals.
3.2 TRANSISTOR CONSTRUCTION
The transistor is a three-layer semiconductor device consisting of either two n- and
one p-type layers of material or two p- and one n-type layers of material. The former
is called an npn transistor, while the latter is called a pnp transistor. Both are shown
in Fig. 3.2 with the proper dc biasing. We will find in Chapter 4 that the dc biasing
is necessary to establish the proper region of operation for ac amplification. The emitter layer is heavily doped, the base lightly doped, and the collector only lightly doped.
The outer layers have widths much greater than the sandwiched p- or n-type material. For the transistors shown in Fig. 3.2 the ratio of the total width to that of the center layer is 0.150/0.001 1501. The doping of the sandwiched layer is also considerably less than that of the outer layers (typically, 101 or less). This lower doping
level decreases the conductivity (increases the resistance) of this material by limiting
the number of “free” carriers.
For the biasing shown in Fig. 3.2 the terminals have been indicated by the capital letters E for emitter, C for collector, and B for base. An appreciation for this choice
of notation will develop when we discuss the basic operation of the transistor. The
abbreviation BJT, from bipolar junction transistor, is often applied to this threeterminal device. The term bipolar reflects the fact that holes and electrons participate
in the injection process into the oppositely polarized material. If only one carrier is
employed (electron or hole), it is considered a unipolar device. The Schottky diode
of Chapter 20 is such a device.
Figure 3.2 Types of transistors:
(a) pnp; (b) npn.
3.3 TRANSISTOR OPERATION
The basic operation of the transistor will now be described using the pnp transistor
of Fig. 3.2a. The operation of the npn transistor is exactly the same if the roles played
by the electron and hole are interchanged. In Fig. 3.3 the pnp transistor has been redrawn without the base-to-collector bias. Note the similarities between this situation
and that of the forward-biased diode in Chapter 1. The depletion region has been reduced in width due to the applied bias, resulting in a heavy flow of majority carriers
from the p- to the n-type material.
Figure 3.3 Forward-biased
junction of a pnp transistor.
3.3 Transistor Operation
113
Let us now remove the base-to-emitter bias of the pnp transistor of Fig. 3.2a as
shown in Fig. 3.4. Consider the similarities between this situation and that of the
reverse-biased diode of Section 1.6. Recall that the flow of majority carriers is zero,
resulting in only a minority-carrier flow, as indicated in Fig. 3.4. In summary, therefore:
One p-n junction of a transistor is reverse biased, while the other is forward
biased.
In Fig. 3.5 both biasing potentials have been applied to a pnp transistor, with the
resulting majority- and minority-carrier flow indicated. Note in Fig. 3.5 the widths of
the depletion regions, indicating clearly which junction is forward-biased and which
is reverse-biased. As indicated in Fig. 3.5, a large number of majority carriers will
diffuse across the forward-biased p-n junction into the n-type material. The question
then is whether these carriers will contribute directly to the base current IB or pass
directly into the p-type material. Since the sandwiched n-type material is very thin
and has a low conductivity, a very small number of these carriers will take this path
of high resistance to the base terminal. The magnitude of the base current is typically
on the order of microamperes as compared to milliamperes for the emitter and collector currents. The larger number of these majority carriers will diffuse across the
reverse-biased junction into the p-type material connected to the collector terminal as
indicated in Fig. 3.5. The reason for the relative ease with which the majority carriers can cross the reverse-biased junction is easily understood if we consider that for
the reverse-biased diode the injected majority carriers will appear as minority carriers in the n-type material. In other words, there has been an injection of minority carriers into the n-type base region material. Combining this with the fact that all the
minority carriers in the depletion region will cross the reverse-biased junction of a
diode accounts for the flow indicated in Fig. 3.5.
Figure 3.4 Reverse-biased junction of a pnp
transistor.
Figure 3.5 Majority and minority
carrier flow of a pnp transistor.
Applying Kirchhoff’s current law to the transistor of Fig. 3.5 as if it were a single node, we obtain
IE IC IB
(3.1)
and find that the emitter current is the sum of the collector and base currents. The
collector current, however, is comprised of two components—the majority and minority carriers as indicated in Fig. 3.5. The minority-current component is called the
leakage current and is given the symbol ICO (IC current with emitter terminal Open).
The collector current, therefore, is determined in total by Eq. (3.2).
IC ICmajority ICOminority
114
Chapter 3
Bipolar Junction Transistors
(3.2)
For general-purpose transistors, IC is measured in milliamperes, while ICO is measured in microamperes or nanoamperes. ICO, like Is for a reverse-biased diode, is temperature sensitive and must be examined carefully when applications of wide temperature ranges are considered. It can severely affect the stability of a system at high
temperature if not considered properly. Improvements in construction techniques have
resulted in significantly lower levels of ICO, to the point where its effect can often be
ignored.
3.4 COMMON-BASE CONFIGURATION
The notation and symbols used in conjunction with the transistor in the majority of
texts and manuals published today are indicated in Fig. 3.6 for the common-base configuration with pnp and npn transistors. The common-base terminology is derived
from the fact that the base is common to both the input and output sides of the configuration. In addition, the base is usually the terminal closest to, or at, ground potential. Throughout this book all current directions will refer to conventional (hole)
flow rather than electron flow. This choice was based primarily on the fact that the
vast amount of literature available at educational and industrial institutions employs
conventional flow and the arrows in all electronic symbols have a direction defined
by this convention. Recall that the arrow in the diode symbol defined the direction of
conduction for conventional current. For the transistor:
The arrow in the graphic symbol defines the direction of emitter current (conventional flow) through the device.
All the current directions appearing in Fig. 3.6 are the actual directions as defined
by the choice of conventional flow. Note in each case that IE IC IB. Note also
that the applied biasing (voltage sources) are such as to establish current in the direction indicated for each branch. That is, compare the direction of IE to the polarity
or VEE for each configuration and the direction of IC to the polarity of VCC.
To fully describe the behavior of a three-terminal device such as the commonbase amplifiers of Fig. 3.6 requires two sets of characteristics—one for the driving
point or input parameters and the other for the output side. The input set for the
common-base amplifier as shown in Fig. 3.7 will relate an input current (IE) to an input voltage (VBE) for various levels of output voltage (VCB).
The output set will relate an output current (IC) to an output voltage (VCB) for various levels of input current (IE) as shown in Fig. 3.8. The output or collector set of
characteristics has three basic regions of interest, as indicated in Fig. 3.8: the active,
Figure 3.6 Notation and symbols used with the common-base
configuration: (a) pnp transistor;
(b) npn transistor.
Figure 3.7 Input or driving
point characteristics for a
common-base silicon transistor
amplifier.
3.4 Common-Base Configuration
115
IC (mA)
Active region (unshaded area)
7 mA
7
6 mA
6
4
3
2
5 mA
Saturation region
5
4 mA
3 mA
2 mA
I E = 1 mA
1
Figure 3.8 Output or collector
characteristics for a common-base
transistor amplifier.
I E = 0 mA
0
−1
0
5
10
15
20
V CB (V)
Cutoff region
cutoff, and saturation regions. The active region is the region normally employed for
linear (undistorted) amplifiers. In particular:
In the active region the collector-base junction is reverse-biased, while the
base-emitter junction is forward-biased.
Figure 3.9 Reverse saturation
current.
The active region is defined by the biasing arrangements of Fig. 3.6. At the lower
end of the active region the emitter current (IE) is zero, the collector current is simply that due to the reverse saturation current ICO, as indicated in Fig. 3.8. The current
ICO is so small (microamperes) in magnitude compared to the vertical scale of IC (milliamperes) that it appears on virtually the same horizontal line as IC 0. The circuit
conditions that exist when IE 0 for the common-base configuration are shown in
Fig. 3.9. The notation most frequently used for ICO on data and specification sheets
is, as indicated in Fig. 3.9, ICBO. Because of improved construction techniques, the
level of ICBO for general-purpose transistors (especially silicon) in the low- and midpower ranges is usually so low that its effect can be ignored. However, for higher
power units ICBO will still appear in the microampere range. In addition, keep in mind
that ICBO, like Is, for the diode (both reverse leakage currents) is temperature sensitive. At higher temperatures the effect of ICBO may become an important factor since
it increases so rapidly with temperature.
Note in Fig. 3.8 that as the emitter current increases above zero, the collector current increases to a magnitude essentially equal to that of the emitter current as determined by the basic transistor-current relations. Note also the almost negligible effect
of VCB on the collector current for the active region. The curves clearly indicate that
a first approximation to the relationship between IE and IC in the active region is given
by
IC IE
(3.3)
As inferred by its name, the cutoff region is defined as that region where the collector current is 0 A, as revealed on Fig. 3.8. In addition:
In the cutoff region the collector-base and base-emitter junctions of a transistor are both reverse-biased.
116
Chapter 3
Bipolar Junction Transistors
The saturation region is defined as that region of the characteristics to the left of
VCB 0 V. The horizontal scale in this region was expanded to clearly show the dramatic change in characteristics in this region. Note the exponential increase in collector current as the voltage VCB increases toward 0 V.
In the saturation region the collector-base and base-emitter junctions are
forward-biased.
The input characteristics of Fig. 3.7 reveal that for fixed values of collector voltage (VCB), as the base-to-emitter voltage increases, the emitter current increases in a
manner that closely resembles the diode characteristics. In fact, increasing levels of
VCB have such a small effect on the characteristics that as a first approximation the
change due to changes in VCB can be ignored and the characteristics drawn as shown
in Fig. 3.10a. If we then apply the piecewise-linear approach, the characteristics of
Fig. 3.10b will result. Taking it a step further and ignoring the slope of the curve and
therefore the resistance associated with the forward-biased junction will result in the
characteristics of Fig. 3.10c. For the analysis to follow in this book the equivalent
model of Fig. 3.10c will be employed for all dc analysis of transistor networks. That
is, once a transistor is in the “on” state, the base-to-emitter voltage will be assumed
to be the following:
VBE 0.7 V
(3.4)
In other words, the effect of variations due to VCB and the slope of the input characteristics will be ignored as we strive to analyze transistor networks in a manner that
will provide a good approximation to the actual response without getting too involved
with parameter variations of less importance.
I E (mA)
I E (mA)
I E (mA)
8
8
7
7
6
6
6
5
5
5
4
4
4
3
3
3
2
2
2
1
1
8
7
Any V CB
0
0.2
0.4
0.6 0.8
(a)
1
VBE (V)
0
1
0.7 V
0.2
0.4
0.6 0.8
1
VBE (V)
0
(b)
0.7 V
0.2
0.4
0.6 0.8
1
VBE (V)
(c)
Figure 3.10 Developing the equivalent model to be employed for the base-toemitter region of an amplifier in the dc mode.
It is important to fully appreciate the statement made by the characteristics of Fig.
3.10c. They specify that with the transistor in the “on” or active state the voltage from
base to emitter will be 0.7 V at any level of emitter current as controlled by the external network. In fact, at the first encounter of any transistor configuration in the dc
mode, one can now immediately specify that the voltage from base to emitter is 0.7 V
if the device is in the active region—a very important conclusion for the dc analysis
to follow.
3.4 Common-Base Configuration
117
EXAMPLE 3.1
(a) Using the characteristics of Fig. 3.8, determine the resulting collector current if
IE 3 mA and VCB 10 V.
(b) Using the characteristics of Fig. 3.8, determine the resulting collector current if
IE remains at 3 mA but VCB is reduced to 2 V.
(c) Using the characteristics of Figs. 3.7 and 3.8, determine VBE if IC 4 mA and
VCB 20 V.
(d) Repeat part (c) using the characteristics of Figs. 3.8 and 3.10c.
Solution
(a) The characteristics clearly indicate that IC IE 3 mA.
(b) The effect of changing VCB is negligible and IC continues to be 3 mA.
(c) From Fig. 3.8, IE IC 4 mA. On Fig. 3.7 the resulting level of VBE is about
0.74 V.
(d) Again from Fig. 3.8, IE IC 4 mA. However, on Fig. 3.10c, VBE is 0.7 V for
any level of emitter current.
Alpha ()
In the dc mode the levels of IC and IE due to the majority carriers are related by a
quantity called alpha and defined by the following equation:
I
dc C
IE
(3.5)
where IC and IE are the levels of current at the point of operation. Even though the
characteristics of Fig. 3.8 would suggest that 1, for practical devices the level of
alpha typically extends from 0.90 to 0.998, with most approaching the high end of
the range. Since alpha is defined solely for the majority carriers, Eq. (3.2) becomes
IC IE ICBO
(3.6)
For the characteristics of Fig. 3.8 when IE 0 mA, IC is therefore equal to ICBO,
but as mentioned earlier, the level of ICBO is usually so small that it is virtually undetectable on the graph of Fig. 3.8. In other words, when IE 0 mA on Fig. 3.8, IC
also appears to be 0 mA for the range of VCB values.
For ac situations where the point of operation moves on the characteristic curve,
an ac alpha is defined by
IC
ac IE
VCB constant
(3.7)
The ac alpha is formally called the common-base, short-circuit, amplification factor,
for reasons that will be more obvious when we examine transistor equivalent circuits
in Chapter 7. For the moment, recognize that Eq. (3.7) specifies that a relatively small
change in collector current is divided by the corresponding change in IE with the
collector-to-base voltage held constant. For most situations the magnitudes of ac and
dc are quite close, permitting the use of the magnitude of one for the other. The use
of an equation such as (3.7) will be demonstrated in Section 3.6.
Biasing
The proper biasing of the common-base configuration in the active region can be determined quickly using the approximation IC IE and assuming for the moment that
118
Chapter 3
Bipolar Junction Transistors
Figure 3.11 Establishing the
proper biasing management for a
common-base pnp transistor in
the active region.
IB 0 A. The result is the configuration of Fig. 3.11 for the pnp transistor. The arrow of the symbol defines the direction of conventional flow for IE IC. The dc supplies are then inserted with a polarity that will support the resulting current direction.
For the npn transistor the polarities will be reversed.
Some students feel that they can remember whether the arrow of the device symbol in pointing in or out by matching the letters of the transistor type with the appropriate letters of the phrases “pointing in” or “not pointing in.” For instance, there
is a match between the letters npn and the italic letters of not pointing in and the letters pnp with pointing in.
3.5 TRANSISTOR AMPLIFYING ACTION
Now that the relationship between IC and IE has been established in Section 3.4, the
basic amplifying action of the transistor can be introduced on a surface level using
the network of Fig. 3.12. The dc biasing does not appear in the figure since our interest will be limited to the ac response. For the common-base configuration the ac
input resistance determined by the characteristics of Fig. 3.7 is quite small and typically varies from 10 to 100 . The output resistance as determined by the curves of
Fig. 3.8 is quite high (the more horizontal the curves the higher the resistance) and
typically varies from 50 k to 1 M (100 k for the transistor of Fig. 3.12). The difference in resistance is due to the forward-biased junction at the input (base to emitter) and the reverse-biased junction at the output (base to collector). Using a common
value of 20 for the input resistance, we find that
V
200 mV
Ii i 10 mA
Ri
20 If we assume for the moment that ac 1 (Ic Ie),
IL Ii 10 mA
VL ILR
and
(10 mA)(5 k)
50 V
pnp
Ii
E
+
V i = 200 mV
C
IL
+
B
Ri
Ro
20 Ω
100 k Ω
R
5 k Ω VL
–
–
Figure 3.12 Basic voltage amplification action of the common-base
configuration.
3.5 Transistor Amplifying Action
119
The voltage amplification is
50 V
VL
Av 250
200 mV
Vi
Typical values of voltage amplification for the common-base configuration vary
from 50 to 300. The current amplification (IC/IE) is always less than 1 for the common-base configuration. This latter characteristic should be obvious since IC IE
and is always less than 1.
The basic amplifying action was produced by transferring a current I from a lowto a high-resistance circuit. The combination of the two terms in italics results in the
label transistor; that is,
transfer resistor → transistor
3.6 COMMON-EMITTER CONFIGURATION
The most frequently encountered transistor configuration appears in Fig. 3.13 for the
pnp and npn transistors. It is called the common-emitter configuration since the emitter is common or reference to both the input and output terminals (in this case common to both the base and collector terminals). Two sets of characteristics are again
necessary to describe fully the behavior of the common-emitter configuration: one for
the input or base-emitter circuit and one for the output or collector-emitter circuit.
Both are shown in Fig. 3.14.
Figure 3.13 Notation and symbols used with the common-emitter configuration: (a) npn transistor; (b) pnp transistor.
The emitter, collector, and base currents are shown in their actual conventional
current direction. Even though the transistor configuration has changed, the current
relations developed earlier for the common-base configuration are still applicable.
That is, IE IC IB and IC IE.
For the common-emitter configuration the output characteristics are a plot of the
output current (IC) versus output voltage (VCE) for a range of values of input current
(IB). The input characteristics are a plot of the input current (IB) versus the input voltage (VBE) for a range of values of output voltage (VCE).
120
Chapter 3
Bipolar Junction Transistors
IC (mA)
8
90 µA
80 µA
7
70 µA
6
I B (µA)
50 µA
(Saturation region) 5
40 µA
90
VCE = 20 V
80
4
70
30 µA
60
3
(Active region)
50
20 µA
40
2
30
10 µA
20
1
I B = 0 µA
0
100
VCE = 1 V
VCE = 10 V
60 µA
VCEsat
10
5
15
20 VCE (V)
10
0
0.2
0.4
0.6
0.8
1.0
VBE (V)
(Cutoff region)
~ β I CBO
I CEO =
(a)
(b)
Figure 3.14 Characteristics of a silicon transistor in the common-emitter configuration: (a) collector characteristics; (b) base characteristics.
Note that on the characteristics of Fig. 3.14 the magnitude of IB is in microamperes, compared to milliamperes of IC. Consider also that the curves of IB are not as
horizontal as those obtained for IE in the common-base configuration, indicating that
the collector-to-emitter voltage will influence the magnitude of the collector current.
The active region for the common-emitter configuration is that portion of the
upper-right quadrant that has the greatest linearity, that is, that region in which the
curves for IB are nearly straight and equally spaced. In Fig. 3.14a this region exists
to the right of the vertical dashed line at VCEsat and above the curve for IB equal to
zero. The region to the left of VCEsat is called the saturation region.
In the active region of a common-emitter amplifier the collector-base junction
is reverse-biased, while the base-emitter junction is forward-biased.
You will recall that these were the same conditions that existed in the active region of the common-base configuration. The active region of the common-emitter
configuration can be employed for voltage, current, or power amplification.
The cutoff region for the common-emitter configuration is not as well defined as
for the common-base configuration. Note on the collector characteristics of Fig. 3.14
that IC is not equal to zero when IB is zero. For the common-base configuration, when
the input current IE was equal to zero, the collector current was equal only to the reverse saturation current ICO, so that the curve IE 0 and the voltage axis were, for
all practical purposes, one.
The reason for this difference in collector characteristics can be derived through
the proper manipulation of Eqs. (3.3) and (3.6). That is,
Eq. (3.6):
Substitution gives
Rearranging yields
Eq. (3.3):
IC IE ICBO
IC (IC IB) ICBO
IB
ICBO
IC 1 1 (3.8)
3.6 Common-Emitter Configuration
121
If we consider the case discussed above, where IB 0 A, and substitute a typical
value of such as 0.996, the resulting collector current is the following:
(0 A)
ICBO
IC 1 1 0.996
ICBO
250ICBO
0.004
If ICBO were 1 A, the resulting collector current with IB 0 A would be
250(1 A) 0.25 mA, as reflected in the characteristics of Fig. 3.14.
For future reference, the collector current defined by the condition IB 0 A will
be assigned the notation indicated by Eq. (3.9).
ICBO
ICEO 1 IB 0 A
(3.9)
In Fig. 3.15 the conditions surrounding this newly defined current are demonstrated
with its assigned reference direction.
For linear (least distortion) amplification purposes, cutoff for the commonemitter configuration will be defined by IC ICEO.
In other words, the region below IB 0 A is to be avoided if an undistorted output signal is required.
When employed as a switch in the logic circuitry of a computer, a transistor will
have two points of operation of interest: one in the cutoff and one in the saturation
region. The cutoff condition should ideally be IC 0 mA for the chosen VCE voltage.
Since ICEO is typically low in magnitude for silicon materials, cutoff will exist for
switching purposes when IB 0 A or IC ICEO for silicon transistors only. For germanium transistors, however, cutoff for switching purposes will be defined as those
conditions that exist when IC ICBO. This condition can normally be obtained for
germanium transistors by reverse-biasing the base-to-emitter junction a few tenths of
a volt.
Recall for the common-base configuration that the input set of characteristics was
approximated by a straight-line equivalent that resulted in VBE 0.7 V for any level
of IE greater than 0 mA. For the common-emitter configuration the same approach
can be taken, resulting in the approximate equivalent of Fig. 3.16. The result supports
our earlier conclusion that for a transistor in the “on” or active region the base-toemitter voltage is 0.7 V. In this case the voltage is fixed for any level of base current.
I B (µA)
100
90
80
70
60
50
40
30
20
10
0
Figure 3.15 Circuit conditions
related to ICEO.
122
0.2
0.4
0.6
0.8
0.7 V
Chapter 3
Bipolar Junction Transistors
1
V BE (V)
Figure 3.16 Piecewise-linear
equivalent for the diode characteristics of Fig. 3.14b.
(a) Using the characteristics of Fig. 3.14, determine IC at IB 30 A and VCE 10 V.
(b) Using the characteristics of Fig. 3.14, determine IC at VBE 0.7 V and VCE 15 V.
EXAMPLE 3.2
Solution
(a) At the intersection of IB 30 A and VCE 10 V, IC 3.4 mA.
(b) Using Fig. 3.14b, IB 20 A at VBE 0.7 V. From Fig. 3.14a we find that IC 2.5 mA at the intersection of IB 20 A and VCE 15 V.
Beta ( )
In the dc mode the levels of IC and IB are related by a quantity called beta and defined by the following equation:
I
C
IB
dc
(3.10)
where IC and IB are determined at a particular operating point on the characteristics.
For practical devices the level of typically ranges from about 50 to over 400, with
most in the midrange. As for , certainly reveals the relative magnitude of one current to the other. For a device with a of 200, the collector current is 200 times the
magnitude of the base current.
On specification sheets dc is usually included as hFE with the h derived from an
ac hybrid equivalent circuit to be introduced in Chapter 7. The subscripts FE are derived from forward-current amplification and common-emitter configuration, respectively.
For ac situations an ac beta has been defined as follows:
ac
IC
IB
VCE constant
(3.11)
The formal name for ac is common-emitter, forward-current, amplification factor.
Since the collector current is usually the output current for a common-emitter configuration and the base current the input current, the term amplification is included
in the nomenclature above.
Equation (3.11) is similar in format to the equation for ac in Section 3.4. The
procedure for obtaining ac from the characteristic curves was not described because
of the difficulty of actually measuring changes of IC and IE on the characteristics.
Equation (3.11), however, is one that can be described with some clarity, and in fact,
the result can be used to find ac using an equation to be derived shortly.
On specification sheets ac is normally referred to as hfe. Note that the only difference between the notation used for the dc beta, specifically, dc hFE, is the type
of lettering for each subscript quantity. The lowercase letter h continues to refer to
the hybrid equivalent circuit to be described in Chapter 7 and the fe to the forward
current gain in the common-emitter configuration.
The use of Eq. (3.11) is best described by a numerical example using an actual
set of characteristics such as appearing in Fig. 3.14a and repeated in Fig. 3.17. Let
us determine ac for a region of the characteristics defined by an operating point of
IB 25 A and VCE 7.5 V as indicated on Fig. 3.17. The restriction of VCE constant requires that a vertical line be drawn through the operating point at VCE 7.5 V.
At any location on this vertical line the voltage VCE is 7.5 V, a constant. The change
3.6 Common-Emitter Configuration
123
I C (mA)
9
90 µA
8
80 µA
7
70 µA
60 µA
6
50 µA
5
40 µA
4
IC2
∆ IC
IC1
30 µA
IB 2
3
25 µA
20 µA
Q - pt.
IB1
2
10 µA
1
IB = 0 µA
0
5
10
15
20
VCE = 7.5 V
Figure 3.17 Determining
ac
and
dc
25
VCE (V)
from the collector characteristics.
in IB (IB) as appearing in Eq. (3.11) is then defined by choosing two points on either side of the Q-point along the vertical axis of about equal distances to either side
of the Q-point. For this situation the IB 20 A and 30 A curves meet the requirement without extending too far from the Q-point. They also define levels of IB
that are easily defined rather than have to interpolate the level of IB between the curves.
It should be mentioned that the best determination is usually made by keeping the
chosen IB as small as possible. At the two intersections of IB and the vertical axis,
the two levels of IC can be determined by drawing a horizontal line over to the vertical axis and reading the resulting values of IC. The resulting ac for the region can
then be determined by
IC2 IC1
IC
ac IB VCE constant
IB2 IB1
1 mA
3.2 mA 2.2 mA
10 A
30 A 20 A
100
The solution above reveals that for an ac input at the base, the collector current will
be about 100 times the magnitude of the base current.
If we determine the dc beta at the Q-point:
dc
124
Chapter 3
I
2.7 mA
C 108
IB
25 A
Bipolar Junction Transistors
Although not exactly equal, the levels of ac and dc are usually reasonably close
and are often used interchangeably. That is, if ac is known, it is assumed to be about
the same magnitude as dc, and vice versa. Keep in mind that in the same lot, the
value of ac will vary somewhat from one transistor to the next even though each
transistor has the same number code. The variation may not be significant but for the
majority of applications, it is certainly sufficient to validate the approximate approach
above. Generally, the smaller the level of ICEO, the closer the magnitude of the two
betas. Since the trend is toward lower and lower levels of ICEO, the validity of the
foregoing approximation is further substantiated.
If the characteristics had the appearance of those appearing in Fig. 3.18, the level
of ac would be the same in every region of the characteristics. Note that the step in
IB is fixed at 10 A and the vertical spacing between curves is the same at every point
in the characteristics—namely, 2 mA. Calculating the ac at the Q-point indicated will
result in
ac
IC
IB
VCE constant
9 mA 7 mA
2 mA
200
45 A 35 A
10 A
Determining the dc beta at the same Q-point will result in
dc
8 mA
I
C 200
40 A
IB
revealing that if the characteristics have the appearance of Fig. 3.18, the magnitude
of ac and dc will be the same at every point on the characteristics. In particular, note
that ICEO 0 A.
Although a true set of transistor characteristics will never have the exact appearance of Fig. 3.18, it does provide a set of characteristics for comparison with those
obtained from a curve tracer (to be described shortly).
Figure 3.18 Characteristics in which
ac
is the same everywhere and
ac
dc.
For the analysis to follow the subscript dc or ac will not be included with to
avoid cluttering the expressions with unnecessary labels. For dc situations it will simply be recognized as dc and for any ac analysis as ac. If a value of is specified
for a particular transistor configuration, it will normally be used for both the dc and
ac calculations.
3.6 Common-Emitter Configuration
125
A relationship can be developed between and using the basic relationships
introduced thus far. Using IC/IB we have IB IC/ , and from IC/IE we have
IE IC/. Substituting into
IE IC IB
I
I
C IC C
we have
and dividing both sides of the equation by IC will result in
1
1
1 ( 1)
or
so that
1
(3.12a)
1 (3.12b)
or
In addition, recall that
ICBO
ICEO 1 but using an equivalence of
1
1 1
derived from the above, we find that
ICEO ( 1)ICBO
or
ICEO ICBO
(3.13)
as indicated on Fig. 3.14a. Beta is a particularly important parameter because it
provides a direct link between current levels of the input and output circuits for a
common-emitter configuration. That is,
IC IB
and since
(31.4)
IE IC IB
IB IB
we have
IE ( 1)IB
(3.15)
Both of the equations above play a major role in the analysis in Chapter 4.
Biasing
The proper biasing of a common-emitter amplifier can be determined in a manner
similar to that introduced for the common-base configuration. Let us assume that we
are presented with an npn transistor such as shown in Fig. 3.19a and asked to apply
the proper biasing to place the device in the active region.
The first step is to indicate the direction of IE as established by the arrow in the
transistor symbol as shown in Fig. 3.19b. Next, the other currents are introduced as
126
Chapter 3
Bipolar Junction Transistors
Figure 3.19 Determining the proper biasing arrangement for a commonemitter npn transistor configuration.
shown, keeping in mind the Kirchhoff’s current law relationship: IC IB IE. Finally, the supplies are introduced with polarities that will support the resulting directions of IB and IC as shown in Fig. 3.19c to complete the picture. The same approach
can be applied to pnp transistors. If the transistor of Fig. 3.19 was a pnp transistor,
all the currents and polarities of Fig. 3.19c would be reversed.
3.7 COMMON-COLLECTOR
CONFIGURATION
The third and final transistor configuration is the common-collector configuration,
shown in Fig. 3.20 with the proper current directions and voltage notation. The
common-collector configuration is used primarily for impedance-matching purposes
since it has a high input impedance and low output impedance, opposite to that of the
common-base and common-emitter configurations.
IE
IE
E
p
IB
V EE
p
p
B
V EE
n
V BB
V BB
C
n
IB
n
B
E
C
IC
IC
IE
IE
E
E
IB
IB
B
B
IC
IC
C
C
(a)
(b)
3.7 Common-Collector Configuration
Figure 3.20 Notation and symbols used with the common-collector configuration: (a) pnp transistor; (b) npn transistor.
127
C
B
E
R
Figure 3.21 Common-collector
configuration used for
impedance-matching purposes.
A common-collector circuit configuration is provided in Fig. 3.21 with the load
resistor connected from emitter to ground. Note that the collector is tied to ground
even though the transistor is connected in a manner similar to the common-emitter
configuration. From a design viewpoint, there is no need for a set of commoncollector characteristics to choose the parameters of the circuit of Fig. 3.21. It can
be designed using the common-emitter characteristics of Section 3.6. For all practical purposes, the output characteristics of the common-collector configuration are the
same as for the common-emitter configuration. For the common-collector configuration the output characteristics are a plot of IE versus VEC for a range of values of IB.
The input current, therefore, is the same for both the common-emitter and commoncollector characteristics. The horizontal voltage axis for the common-collector configuration is obtained by simply changing the sign of the collector-to-emitter voltage
of the common-emitter characteristics. Finally, there is an almost unnoticeable change
in the vertical scale of IC of the common-emitter characteristics if IC is replaced by
IE for the common-collector characteristics (since 1). For the input circuit of the
common-collector configuration the common-emitter base characteristics are sufficient for obtaining the required information.
3.8 LIMITS OF OPERATION
For each transistor there is a region of operation on the characteristics which will ensure that the maximum ratings are not being exceeded and the output signal exhibits
minimum distortion. Such a region has been defined for the transistor characteristics
of Fig. 3.22. All of the limits of operation are defined on a typical transistor specification sheet described in Section 3.9.
Some of the limits of operation are self-explanatory, such as maximum collector
current (normally referred to on the specification sheet as continuous collector current) and maximum collector-to-emitter voltage (often abbreviated as VCEO or V(BR)CEO
on the specification sheet). For the transistor of Fig. 3.22, ICmax was specified as 50 mA
and VCEO as 20 V. The vertical line on the characteristics defined as VCEsat specifies
Figure 3.22 Defining the linear
(undistorted) region of operation
for a transistor.
128
Chapter 3
Bipolar Junction Transistors
the minimum VCE that can be applied without falling into the nonlinear region labeled
the saturation region. The level of VCEsat is typically in the neighborhood of the 0.3 V
specified for this transistor.
The maximum dissipation level is defined by the following equation:
PCmax VCEIC
(3.16)
For the device of Fig. 3.22, the collector power dissipation was specified as
300 mW. The question then arises of how to plot the collector power dissipation curve
specified by the fact that
PCmax VCEIC 300 mW
or
VCEIC 300 mW
At any point on the characteristics the product of VCE and IC must be equal to
300 mW. If we choose IC to be the maximum value of 50 mA and substitute into the
relationship above, we obtain
VCEIC 300 mW
VCE (50 mA) 300 mW
300 mW
VCE 6 V
50 mA
As a result we find that if IC 50 mA, then VCE 6 V on the power dissipation
curve as indicated in Fig. 3.22. If we now choose VCE to be its maximum value of
20 V, the level of IC is the following:
(20 V)IC 300 mW
300 mW
IC 15 mA
20 V
defining a second point on the power curve.
If we now choose a level of IC in the midrange such as 25 mA, and solve for the
resulting level of VCE, we obtain
VCE(25 mA) 300 mW
and
300 mW
VCE 12 V
25 mA
as also indicated on Fig. 3.22.
A rough estimate of the actual curve can usually be drawn using the three points
defined above. Of course, the more points you have, the more accurate the curve, but
a rough estimate is normally all that is required.
The cutoff region is defined as that region below IC ICEO. This region must also
be avoided if the output signal is to have minimum distortion. On some specification
sheets only ICBO is provided. One must then use the equation ICEO ICBO to establish some idea of the cutoff level if the characteristic curves are unavailable. Operation in the resulting region of Fig. 3.22 will ensure minimum distortion of the output signal and current and voltage levels that will not damage the device.
If the characteristic curves are unavailable or do not appear on the specification
sheet (as is often the case), one must simply be sure that IC, VCE, and their product
VCEIC fall into the range appearing in Eq. (3.17).
3.8 Limits of Operation
129
ICEO
VCEsat
IC
VCE
VCEIC
ICmax
VCEmax
(3.17)
PCmax
For the common-base characteristics the maximum power curve is defined by the following product of output quantities:
PCmax VCBIC
(3.18)
3.9 TRANSISTOR SPECIFICATION SHEET
Since the specification sheet is the communication link between the manufacturer and
user, it is particularly important that the information provided be recognized and correctly understood. Although all the parameters have not been introduced, a broad number will now be familiar. The remaining parameters will be introduced in the chapters that follow. Reference will then be made to this specification sheet to review the
manner in which the parameter is presented.
The information provided as Fig. 3.23 is taken directly from the Small-Signal
Transistors, FETs, and Diodes publication prepared by Motorola Inc. The 2N4123 is
a general-purpose npn transistor with the casing and terminal identification appearing in the top-right corner of Fig. 3.23a. Most specification sheets are broken down
into maximum ratings, thermal characteristics, and electrical characteristics. The
electrical characteristics are further broken down into “on,” “off,” and small-signal
characteristics. The “on” and “off” characteristics refer to dc limits, while the smallsignal characteristics include the parameters of importance to ac operation.
Note in the maximum rating list that VCEmax VCEO 30 V with ICmax 200 mA.
The maximum collector dissipation PCmax PD 625 mW. The derating factor under the maximum rating specifies that the maximum rating must be decreased 5 mW
for every 1° rise in temperature above 25°C. In the “off” characteristics ICBO is specified as 50 nA and in the “on” characteristics VCEsat 0.3 V. The level of hFE has a
range of 50 to 150 at IC 2 mA and VCE 1 V and a minimum value of 25 at a
higher current of 50 mA at the same voltage.
The limits of operation have now been defined for the device and are repeated below in the format of Eq. (3.17) using hFE 150 (the upper limit) and ICEO ICBO (150)(50 nA) 7.5 A. Certainly, for many applications the 7.5 A 0.0075 mA
can be considered to be 0 mA on an approximate basis.
Limits of Operation
7.5 mA IC 200 mA
0.3 V
VCE
VCEIC
30 V
650 mW
In the small-signal characteristics the level of hfe ( ac) is provided along with a
plot of how it varies with collector current in Fig. 3.23f. In Fig. 3.23j the effect of
temperature and collector current on the level of hFE ( ac) is demonstrated. At room
temperature (25°C), note that hFE ( dc) is a maximum value of 1 in the neighborhood
of about 8 mA. As IC increased beyond this level, hFE drops off to one-half the value
with IC equal to 50 mA. It also drops to this level if IC decreases to the low level of
0.15 mA. Since this is a normalized curve, if we have a transistor with dc hFE 50 at room temperature, the maximum value at 8 mA is 50. At IC 50 mA it has
dropped to 50/2 25. In other words, normalizing reveals that the actual level of hFE
130
Chapter 3
Bipolar Junction Transistors
at any level of IC has been divided by the maximum value of hFE at that temperature
and IC 8 mA. Note also that the horizontal scale of Fig. 3.23j is a log scale. Log
scales are examined in depth in Chapter 11. You may want to look back at the plots
of this section when you find time to review the first few sections of Chapter 11.
Figure 3.23 Transistor specification sheet.
3.9 Transistor Specification Sheet
131
Before leaving this description of the characteristics, take note of the fact that the
actual collector characteristics are not provided. In fact, most specification sheets as
provided by the range of manufacturers fail to provide the full characteristics. It is
expected that the data provided are sufficient to use the device effectively in the design process.
As noted in the introduction to this section, all the parameters of the specification
sheet have not been defined in the preceding sections or chapters. However, the specification sheet provided in Fig. 3.23 will be referenced continually in the chapters to
follow as parameters are introduced. The specification sheet can be a very valuable
tool in the design or analysis mode, and every effort should be made to be aware of
the importance of each parameter and how it may vary with changing levels of current, temperature, and so on.
Figure 1 – Capacitance
Figure 2 – Switching Times
10
200
ts
100
5.0
70
50
C ibo
Time (ns)
Capacitance (pF)
7.0
3.0
20
Cobo
2.0
10.0
7.0
1.0
0.1
0.2 0.3 0.5 0.7 1.0
2.0 3.0 5.0 7.0 10
Reverse bias voltage (V)
20 30 40
td
tr
30
tf
VCC = 3 V
IC / IB = 10
VEB (off) = 0.5 V
5.0
1.0
2.0 3.0
5.0
10
20 30 50
I C , Collector current (mA)
(b)
100
200
(c)
AUDIO SMALL SIGNAL CHARACTERISTICS
NOISE FIGURE
(VCE = 5 Vdc, TA = 25°C)
Bandwidth = 1.0 Hz
Figure 3 – Frequency Variations
Figure 4 – Source Resistance
12
14
f = 1 kHz
Source resistance = 200 Ω
IC = 1 mA
Source resistance = 200 Ω
IC = 0.5 mA
8
MF, Noise figure (dB)
MF, Noise figure (dB)
10
12
Source resistance = 1 k Ω
IC = 50 µ A
6
4
2
0
0.1
0.4
1
8
IC = 50 µ A
6
4
2
4
10
f, Frequency (kHz)
20
40
100
0
0.1
(d)
IC = 100 µ A
0.2
0.4
1.0 2.0 4.0
10 20
R S , Source Resistance (kΩ )
(e)
Figure 3.23 Continued.
132
IC = 0.5 mA
2
Source resistance = 500 Ω
IC = 100 µ A
0.2
IC = 1 mA
10
Chapter 3
Bipolar Junction Transistors
40
100
h PARAMETERS
VCE = 10 V, f = 1 kHz, TA = 25°C
Figure 5 – Current Gain
Figure 6 – Output Admittance
100
h oe Output admittance (µ mhos)
300
h fe Current gain
200
100
70
50
30
0.1
0.2
0.5
1.0
2.0
I C , Collector current (mA)
5.0
50
20
10
5.0
2.0
1.0
0.1
10
0.2
0.5
1.0
2.0
I C , Collector current (mA)
(f)
10
(g)
Figure 7 – Input Impedance
Figure 8 – Voltage Feedback Ratio
10
h re Voltage feedback ratio (× 10−4 )
20
h ie Input impedance (kΩ)
5.0
10
5.0
2.0
1.0
0.5
0.2
0.1
0.2
0.5
1.0
2.0
I C , Collector current (mA)
5.0
(h)
10
7.0
5.0
3.0
2.0
1.0
0.7
0.5
0.1
0.2
0.5
1.0
2.0
I C , Collector current (mA)
10
(i)
STATIC CHARACTERISTICS
(h)
5.0
(i)
STATIC CHARACTERISTICS
Figure 9 – DC Current Gain
h FE DC Current gain (normalized)
2.0
VCE = 1 V
TJ = +125° C
+25° C
1.0
0.7
–55°C
0.5
0.3
0.2
0.1
0.1
0.2 0.3
0.5 0.7 1.0
2.0 3.0
5.0 7.0 10
I C , Collector current (mA)
20
30
50
70 100
200
(j)
Figure 3.23 Continued.
3.9 Transistor Specification Sheet
133
3.10 TRANSISTOR TESTING
As with diodes, there are three routes one can take to check a transistor: curve tracer,
digital meter, and ohmmeter.
Curve Tracer
The curve tracer of Fig. 1.45 will provide the display of Fig. 3.24 once all the controls have been properly set. The smaller displays to the right reveal the scaling to be
applied to the characteristics. The vertical sensitivity is 2 mA/div, resulting in the scale
shown to the left of the monitor’s display. The horizontal sensitivity is 1 V/div, resulting in the scale shown below the characteristics. The step function reveals that the
curves are separated by a difference of 10 A, starting at 0 A for the bottom curve.
The last scale factor provided can be used to quickly determine the ac for any region of the characteristics. Simply multiply the displayed factor by the number of divisions between IB curves in the region of interest. For instance, let us determine ac
at a Q-point of IC 7 mA and VCE 5 V. In this region of the display, the distance
between IB curves is 190 of a division, as indicated on Fig. 3.25. Using the factor specified, we find that
ac
9
200
div 180
10
div
20 mA
Vertical
per div
2 mA
18 mA
80 µ A
16 mA
70 µA
14 mA
Horizontal
per div
1V
60 µA
12 mA
50 µA
10 mA
40 µA
Per Step
10 µ A
8 mA
30 µA
6 mA
20 µA
4 mA
B or gm
per div
200
10 µA
2 mA
Figure 3.24 Curve tracer
response to 2N3904 npn
transistor.
0 µA
0 mA
0V
1V
2V
3V
4V
5V
6V
7V
8V
IC = 8 mA
9V
10 V
IB 2 = 40 µ A
IC 2 = 8.2 mA
∆ IC
Figure 3.25 Determining ac
for the transistor characteristics of
Fig. 3.24 at IC 7 mA and
VCE 5 V.
134
9 div
≅ 10
IC 1 = 6.4 mA
IC = 6 mA
Chapter 3
Bipolar Junction Transistors
Q-point
( IC = 7 m A, VCE = 5 V)
IB 1 = 30 µ A
VCE = 5 V
Using Eq. (3.11) gives us
ac
IC
IB
IC2 IC1
8.2 mA 6.4 mA
40 A 30 A
I
I
VCE constant
B2
B1
1.8 mA
180
10 A
verifying the determination above.
Advanced Digital Meters
Advanced digital meters such as that shown in Fig. 3.26 are now available that can
provide the level of hFE using the lead sockets appearing at the bottom left of the dial.
Note the choice of pnp or npn and the availability of two emitter connections to handle the sequence of leads as connected to the casing. The level of hFE is determined
at a collector current of 2 mA for the Testmate 175A, which is also provided on the
digital display. Note that this versatile instrument can also check a diode. It can measure capacitance and frequency in addition to the normal functions of voltage, current, and resistance measurements.
In fact, in the diode testing mode it can be used to check the p-n junctions of a
transistor. With the collector open the base-to-emitter junction should result in a low
voltage of about 0.7 V with the red (positive) lead connected to the base and the black
(negative) lead connected to the emitter. A reversal of the leads should result in an
OL indication to represent the reverse-biased junction. Similarly, with the emitter
open, the forward- and reverse-bias states of the base-to-collector junction can be
checked.
Figure 3.26 Transistor tester.
(Courtesy Computronics Technology, Inc.)
Ohmmeter
An ohmmeter or the resistance scales of a DMM can be used to check the state of a
transistor. Recall that for a transistor in the active region the base-to-emitter junction
is forward-biased and the base-to-collector junction is reverse-biased. Essentially,
therefore, the forward-biased junction should register a relatively low resistance while
the reverse-biased junction shows a much higher resistance. For an npn transistor, the
forward-biased junction (biased by the internal supply in the resistance mode) from
base to emitter should be checked as shown in Fig. 3.27 and result in a reading that
will typically fall in the range of 100 to a few kilohms. The reverse-biased baseto-collector junction (again reverse-biased by the internal supply) should be checked
as shown in Fig. 3.28 with a reading typically exceeding 100 k. For a pnp transistor the leads are reversed for each junction. Obviously, a large or small resistance in
both directions (reversing the leads) for either junction of an npn or pnp transistor indicates a faulty device.
If both junctions of a transistor result in the expected readings the type of transistor can also be determined by simply noting the polarity of the leads as applied to
the base-emitter junction. If the positive () lead is connected to the base and the
negative lead ( ) to the emitter a low resistance reading would indicate an npn transistor. A high resistance reading would indicate a pnp transistor. Although an ohmmeter can also be used to determine the leads (base, collector and emitter) of a transistor it is assumed that this determination can be made by simply looking at the
orientation of the leads on the casing.
3.10 Transistor Testing
Low R
Open
Ω
+ –
B
E
Figure 3.27 Checking the
forward-biased base-to-emitter
junction of an npn transistor.
High R
Ω
+ –
C
B
E
Figure 3.28 Checking the
reverse-biased base-to-collector
junction of an npn transistor.
135
3.11 TRANSISTOR CASING AND
TERMINAL IDENTIFICATION
After the transistor has been manufactured using one of the techniques described in
Chapter 12, leads of, typically, gold, aluminum, or nickel are then attached and the
entire structure is encapsulated in a container such as that shown in Fig. 3.29. Those
with the heavy duty construction are high-power devices, while those with the small
can (top hat) or plastic body are low- to medium-power devices.
Figure 3.29 Various types of transistors: (a) Courtesy General Electric Company;
(b) and (c) Courtesy of Motorola Inc.; (d) Courtesy International Rectifier Corporation.
Whenever possible, the transistor casing will have some marking to indicate which
leads are connected to the emitter, collector, or base of a transistor. A few of the methods commonly used are indicated in Fig. 3.30.
Figure 3.30 Transistor terminal identification.
The internal construction of a TO-92 package in the Fairchild line appears in Fig.
3.31. Note the very small size of the actual semiconductor device. There are gold
bond wires, a copper frame, and an epoxy encapsulation.
Four (quad) individual pnp silicon transistors can be housed in the 14-pin plastic
dual-in-line package appearing in Fig. 3.32a. The internal pin connections appear in
Fig. 3.32b. As with the diode IC package, the indentation in the top surface reveals
the number 1 and 14 pins.
136
Chapter 3
Bipolar Junction Transistors
Figure 3.31 Internal construction of a Fairchild transistor in a TO-92 package.
(Courtesy Fairchild Camera and Instrument Corporation.)
(Top View)
C
B
E
NC
E
B
C
14
13
12
11
10
9
8
1
2
3
4
5
6
7
C
B
E
NC
E
B
C
NC – No internal connection
(a)
(b)
Figure 3.32 Type Q2T2905 Texas Instruments quad pnp silicon transistors:
(a) appearance; (b) pin connections. (Courtesy Texas Instruments Incorporated.)
3.11 Transistor Casing and Terminal Identification
137
3.12 PSPICE WINDOWS
Since the transistor characteristics were introduced in this chapter it seems appropriate that a procedure for obtaining those characteristics using PSpice Windows should
be examined. The transistors are listed in the EVAL.slb library and start with the letter Q. The library includes two npn transistors and two pnp transistors. The fact that
there are a series of curves defined by the levels of IB will require that a sweep of IB
values (a nested sweep) occur within a sweep of collector-to-emitter voltages. This is
unnecessary for the diode, however, since only one curve would result.
First, the network in Fig. 3.33 is established using the same procedure defined in
Chapter 2. The voltage VCC will establish our main sweep while the voltage VBB will
determine the nested sweep. For future reference, note the panel at the top right of
the menu bar with the scroll control when building networks. This option allows you
to retrieve elements that have been used in the past. For instance, if you placed a resistor a few elements ago, simply return to the scroll bar and scroll until the resistor
R appears. Click the location once, and the resistor will appear on the screen.
Figure 3.33 Network employed to obtain the collector characteristics of the
Q2N2222 transistor.
Next, choose the Analysis Setup icon and enable the DC Sweep. Click on DC
Sweep, and choose Voltage Source and Linear. Type in the Name VCC with a Start
Value of 0 V and an End Value of 10 V. Use an Increment of 0.01 V to ensure a continuous, detailed plot. Rather than click OK, this time we have to choose the Nested
Sweep at the bottom left of the dialog box. When chosen, a DC Nested Sweep dialog
box will appear and ask us to repeat the choices just made for the voltage VBB. Again,
Voltage Source and Linear are chosen, and the name is inserted as VBB. The Start Value
will now be 2.7 V to correspond with an initial current of 20 A as determined by
VBB VBE
2.7 V 0.7 V
20 A
IB RB
100 k
The Increment will be 2V, corresponding with a change in base current of 20 A
between IB levels. The final value will be 10.7 V, corresponding with a current of 100
A. Before leaving the dialog box, be sure to enable the nested sweep. Then, choose
OK, followed by a closing of the Analysis Setup, and we are ready for the analysis.
This time we will automatically Run Probe after the analysis by choosing AnalysisProbe Setup, followed by selecting Automatically run Probe after simulation. After choosing OK, followed by a clicking of the Simulation icon (recall that it was the
138
Chapter 3
Bipolar Junction Transistors
icon with the yellow background and two waveforms), the OrCAD MicroSim Probe
screen will automatically appear. This time, since VCC is the collector-to-emitter voltage, there is no need to label the voltage at the collector. In fact, since it appears as
the horizontal axis of the Probe response, there is no need to touch the X-Axis Settings at all if we recognize that VCC is the collector-to-emitter voltage. For the vertical axis, we turn to Trace-Add and obtain the Add Traces dialog box. Choosing
IC(Q1) and OK, we obtain the transistor characteristics. Unfortunately, however, they
extend from 10 to 20 mA on the vertical axis. This can be corrected by choosing
Plot and then Y-Axis Settings to obtain the Y-Axis Settings dialog box. By choosing User Defined, the range can be set from 0 to 20 mA with a Linear scale. Choose
OK again, and the characteristics of Fig. 3.34 result.
Figure 3.34 Collector characteristics for the transistor of Figure 3.33.
Using the ABC icon on the menu bar, the various levels of IB can be inserted
along with the axis labels VCE and IC. Simply click on the icon, and a dialog box appears asking for the text material. Enter the desired text, click OK, and it will appear
on the screen. It can then be placed in the desired location.
If the ac beta is determined in the middle of the graph, you will find that its value is
about 190—even though Bf in the list of specifications is 255.9. Again, like the diode,
the other parameters of the element have a noticeable effect on the total operation.
However, if we return to the diode specifications through Edit-Model-Edit
Instance Model (Text) and remove all parameters of the device except Bf 255.9 (don’t
forget the close parentheses at the end of the listing) and follow with an OK and a Simulation, a new set of curves will result. An adjustment of the range of the y-axis to
0–30 mA using the Y-Axis Settings will result in the characteristic curves of Fig. 3.35.
Note first that the curves are all horizontal, meaning that the element is void of
any resistive elements. In addition, the equal spacing of the curves throughout reveals
that beta is the same everywhere (as specified by our new device characteristics). Using a difference of 5 mA between any two curves and dividing by the difference in
IB of 20 A will result in a of 250, which is essentially the same as that specified
for the device.
3.12 PSpice Windows
139
FIGURE 3.35 Ideal collector
characteristics for the transistor of
Figure 3.33.
PROBLEMS
§ 3.2 Transistor Construction
1. What names are applied to the two types of BJT transistors? Sketch the basic construction of
each and label the various minority and majority carriers in each. Draw the graphic symbol next
to each. Is any of this information altered by changing from a silicon to a germanium base?
2. What is the major difference between a bipolar and a unipolar device?
§ 3.3 Transistor Operation
3. How must the two transistor junctions be biased for proper transistor amplifier operation?
4. What is the source of the leakage current in a transistor?
5. Sketch a figure similar to Fig. 3.3 for the forward-biased junction of an npn transistor. Describe
the resulting carrier motion.
6. Sketch a figure similar to Fig. 3.4 for the reverse-biased junction of an npn transistor. Describe
the resulting carrier motion.
7. Sketch a figure similar to Fig. 3.5 for the majority- and minority-carrier flow of an npn transistor. Describe the resulting carrier motion.
8. Which of the transistor currents is always the largest? Which is always the smallest? Which
two currents are relatively close in magnitude?
9. If the emitter current of a transistor is 8 mA and IB is 1/100 of IC, determine the levels of IC
and IB.
§ 3.4 Common-Base Configuration
10. From memory, sketch the transistor symbol for a pnp and an npn transistor, and then insert the
conventional flow direction for each current.
11. Using the characteristics of Fig. 3.7, determine VBE at IE 5 mA for VCB 1, 10, and 20 V.
Is it reasonable to assume on an approximate basis that VCB has only a slight effect on the relationship between VBE and IE?
12. (a) Determine the average ac resistance for the characteristics of Fig. 3.10b.
(b) For networks in which the magnitude of the resistive elements is typically in kilohms, is
the approximation of Fig. 3.10c a valid one [based on the results of part (a)]?
140
Chapter 3
Bipolar Junction Transistors
13. (a) Using the characteristics of Fig. 3.8, determine the resulting collector current if IE 4.5 mA
and VCB 4 V.
(b) Repeat part (a) for IE 4.5 mA and VCB 16 V.
(c) How have the changes in VCB affected the resulting level of IC?
(d) On an approximate basis, how are IE and IC related based on the results above?
14. (a) Using the characteristics of Figs. 3.7 and 3.8, determine IC if VCB 10 V and VBE 800 mV.
(b) Determine VBE if IC 5 mA and VCB 10 V.
(c) Repeat part (b) using the characteristics of Fig. 3.10b.
(d) Repeat part (b) using the characteristics of Fig. 3.10c.
(e) Compare the solutions for VBE for parts (b), (c), and (d). Can the difference be ignored if
voltage levels greater than a few volts are typically encountered?
15. (a) Given an dc of 0.998, determine IC if IE 4 mA.
(b) Determine dc if IE 2.8 mA and IB 20 A.
(c) Find IE if IB 40 A and dc is 0.98.
16. From memory, and memory only, sketch the common-base BJT transistor configuration (for
npn and pnp) and indicate the polarity of the applied bias and resulting current directions.
§ 3.5 Transistor Amplifying Action
17. Calculate the voltage gain (Av VL/Vi) for the network of Fig. 3.12 if Vi 500 mV and
R 1 k. (The other circuit values remain the same.)
18. Calculate the voltage gain (Av VL/Vi) for the network of Fig. 3.12 if the source has an internal resistance of 100 in series with Vi.
§ 3.6 Common-Emitter Configuration
19. Define ICBO and ICEO. How are they different? How are they related? Are they typically close
in magnitude?
20. Using the characteristics of Fig. 3.14:
(a) Find the value of IC corresponding to VBE 750 mV and VCE 5 V.
(b) Find the value of VCE and VBE corresponding to IC 3 mA and IB 30 A.
* 21. (a) For the common-emitter characteristics of Fig. 3.14, find the dc beta at an operating point
of VCE 8 V and IC 2 mA.
(b) Find the value of corresponding to this operating point.
(c) At VCE 8 V, find the corresponding value of ICEO.
(d) Calculate the approximate value of ICBO using the dc beta value obtained in part (a).
* 22. (a) Using the characteristics of Fig. 3.14a, determine ICEO at VCE 10 V.
(b) Determine dc at IB 10 A and VCE 10 V.
(c) Using the dc determined in part (b), calculate ICBO.
23. (a)
(b)
(c)
(d)
Using the characteristics of Fig. 3.14a, determine dc at IB 80 A and VCE 5 V.
Repeat part (a) at IB 5 A and VCE 15 V.
Repeat part (a) at IB 30 A and VCE 10 V.
Reviewing the results of parts (a) through (c), does the value of dc change from point to
point on the characteristics? Where were the higher values found? Can you develop any
general conclusions about the value of dc on a set of characteristics such as those provided in Fig. 3.14a?
Using the characteristics of Fig. 3.14a, determine ac at IB 80 A and VCE 5 V.
Repeat part (a) at IB 5 A and VCE 15 V.
Repeat part (a) at IB 30 A and VCE 10 V.
Reviewing the results of parts (a) through (c), does the value of ac change from point to
point on the characteristics? Where are the high values located? Can you develop any general conclusions about the value of ac on a set of collector characteristics?
(e) The chosen points in this exercise are the same as those employed in Problem 23. If Problem 23 was performed, compare the levels of dc and ac for each point and comment on
the trend in magnitude for each quantity.
* 24. (a)
(b)
(c)
(d)
Problems
141
25. Using the characteristics of Fig. 3.14a, determine dc at IB 25 A and VCE 10 V. Then calculate dc and the resulting level of IE. (Use the level of IC determined by IC dcIB.)
26. (a) Given that dc 0.987, determine the corresponding value of
(b) Given dc 120, determine the corresponding value of .
(c) Given that dc 180 and IC 2.0 mA, find IE and IB.
dc.
27. From memory, and memory only, sketch the common-emitter configuration (for npn and pnp)
and insert the proper biasing arrangement with the resulting current directions for IB, IC, and
IE.
§ 3.7 Common-Collector Configuration
28. An input voltage of 2 V rms (measured from base to ground) is applied to the circuit of Fig.
3.21. Assuming that the emitter voltage follows the base voltage exactly and that Vbe (rms) 0.1 V, calculate the circuit voltage amplification (Av Vo /Vi ) and emitter current for RE 1 k.
29. For a transistor having the characteristics of Fig. 3.14, sketch the input and output characteristics of the common-collector configuration.
§ 3.8 Limits of Operation
30. Determine the region of operation for a transistor having the characteristics of Fig. 3.14 if
ICmax 7 mA, VCEmax 17 V, and PCmax 40 mW.
31. Determine the region of operation for a transistor having the characteristics of Fig. 3.8 if
ICmax 6 mA, VCBmax 15 V, and PCmax 30 mW.
§ 3.9 Transistor Specification Sheet
32. Referring to Fig. 3.23, determine the temperature range for the device in degrees Fahrenheit.
33. Using the information provided in Fig. 3.23 regarding PDmax, VCEmax, ICmax and VCEsat, sketch the
boundaries of operation for the device.
34. Based on the data of Fig. 3.23, what is the expected value of ICEO using the average value of
dc?
35. How does the range of hFE (Fig. 3.23(j), normalized from hFE 100) compare with the range
of hfe (Fig. 3.23(f)) for the range of IC from 0.1 to 10 mA?
36. Using the characteristics of Fig. 3.23b, determine whether the input capacitance in the
common-base configuration increases or decreases with increasing levels of reverse-bias potential. Can you explain why?
* 37. Using the characteristics of Fig. 3.23f, determine how much the level of hfe has changed from
its value at 1 mA to its value at 10 mA. Note that the vertical scale is a log scale that may require reference to Section 11.2. Is the change one that should be considered in a design situation?
* 38. Using the characteristics of Fig. 3.23j, determine the level of dc at IC 10 mA at the three
levels of temperature appearing in the figure. Is the change significant for the specified temperature range? Is it an element to be concerned about in the design process?
§ 3.10 Transistor Testing
39. (a)
(b)
(c)
(d)
(e)
(f)
Using the characteristics of Fig. 3.24, determine ac at IC 14 mA and VCE 3 V.
Determine dc at IC 1 mA and VCE 8 V.
Determine ac at IC 14 mA and VCE 3 V.
Determine dc at IC 1 mA and VCE 8 V.
How does the level of ac and dc compare in each region?
Is the approximation dc ac a valid one for this set of characteristics?
*Please Note: Asterisks indicate more difficult problems.
142
Chapter 3
Bipolar Junction Transistors
CHAPTER
DC Biasing—BJTs
4
4.1 INTRODUCTION
The analysis or design of a transistor amplifier requires a knowledge of both the dc
and ac response of the system. Too often it is assumed that the transistor is a magical device that can raise the level of the applied ac input without the assistance of an
external energy source. In actuality, the improved output ac power level is the result
of a transfer of energy from the applied dc supplies. The analysis or design of any
electronic amplifier therefore has two components: the dc portion and the ac portion.
Fortunately, the superposition theorem is applicable and the investigation of the dc
conditions can be totally separated from the ac response. However, one must keep in
mind that during the design or synthesis stage the choice of parameters for the required dc levels will affect the ac response, and vice versa.
The dc level of operation of a transistor is controlled by a number of factors, including the range of possible operating points on the device characteristics. In Section 4.2 we specify the range for the BJT amplifier. Once the desired dc current and
voltage levels have been defined, a network must be constructed that will establish
the desired operating point—a number of these networks are analyzed in this chapter. Each design will also determine the stability of the system, that is, how sensitive
the system is to temperature variations—another topic to be investigated in a later
section of this chapter.
Although a number of networks are analyzed in this chapter, there is an underlying similarity between the analysis of each configuration due to the recurring use of
the following important basic relationships for a transistor:
VBE 0.7 V
(4.1)
IE ( 1)IB IC
(4.2)
IC IB
(4.3)
In fact, once the analysis of the first few networks is clearly understood, the path
toward the solution of the networks to follow will begin to become quite apparent. In
most instances the base current IB is the first quantity to be determined. Once IB is
known, the relationships of Eqs. (4.1) through (4.3) can be applied to find the remaining quantities of interest. The similarities in analysis will be immediately obvious as we progress through the chapter. The equations for IB are so similar for a num-
143
ber of configurations that one equation can be derived from another simply by dropping or adding a term or two. The primary function of this chapter is to develop a
level of familiarity with the BJT transistor that would permit a dc analysis of any system that might employ the BJT amplifier.
4.2 OPERATING POINT
The term biasing appearing in the title of this chapter is an all-inclusive term for the
application of dc voltages to establish a fixed level of current and voltage. For transistor amplifiers the resulting dc current and voltage establish an operating point on
the characteristics that define the region that will be employed for amplification of
the applied signal. Since the operating point is a fixed point on the characteristics, it
is also called the quiescent point (abbreviated Q-point). By definition, quiescent means
quiet, still, inactive. Figure 4.1 shows a general output device characteristic with four
operating points indicated. The biasing circuit can be designed to set the device operation at any of these points or others within the active region. The maximum ratings are indicated on the characteristics of Fig. 4.1 by a horizontal line for the maximum collector current ICmax and a vertical line at the maximum collector-to-emitter
voltage VCEmax. The maximum power constraint is defined by the curve PCmax in the
same figure. At the lower end of the scales are the cutoff region, defined by IB 0 , and the saturation region, defined by VCE VCEsat.
The BJT device could be biased to operate outside these maximum limits, but the
result of such operation would be either a considerable shortening of the lifetime of
the device or destruction of the device. Confining ourselves to the active region, one
can select many different operating areas or points. The chosen Q-point often depends
on the intended use of the circuit. Still, we can consider some differences among the
IC (mA)
80 µA
70 µA
IC max 25
60 µA
50 µA
20
40 µA
PC max
15
30 µA
Saturation
B
10
20 µA
D
10 µA
5
C
I B = 0 µA
A
0
VCE sat
5
10
15
20
Cutoff
VCE max
Figure 4.1 Various operating points within the limits of operation of a transistor.
144
Chapter 4
DC Biasing—BJTs
VCE (V)
various points shown in Fig. 4.1 to present some basic ideas about the operating point
and, thereby, the bias circuit.
If no bias were used, the device would initially be completely off, resulting in a
Q-point at A—namely, zero current through the device (and zero voltage across it).
Since it is necessary to bias a device so that it can respond to the entire range of an
input signal, point A would not be suitable. For point B, if a signal is applied to the
circuit, the device will vary in current and voltage from operating point, allowing the
device to react to (and possibly amplify) both the positive and negative excursions of
the input signal. If the input signal is properly chosen, the voltage and current of the
device will vary but not enough to drive the device into cutoff or saturation. Point C
would allow some positive and negative variation of the output signal, but the peakto-peak value would be limited by the proximity of VCE 0V/IC 0 mA. Operating
at point C also raises some concern about the nonlinearities introduced by the fact
that the spacing between IB curves is rapidly changing in this region. In general, it is
preferable to operate where the gain of the device is fairly constant (or linear) to ensure that the amplification over the entire swing of input signal is the same. Point B
is a region of more linear spacing and therefore more linear operation, as shown in
Fig. 4.1. Point D sets the device operating point near the maximum voltage and power
level. The output voltage swing in the positive direction is thus limited if the maximum voltage is not to be exceeded. Point B therefore seems the best operating point
in terms of linear gain and largest possible voltage and current swing. This is usually
the desired condition for small-signal amplifiers (Chapter 8) but not the case necessarily for power amplifiers, which will be considered in Chapter 16. In this discussion, we will be concentrating primarily on biasing the transistor for small-signal amplification operation.
One other very important biasing factor must be considered. Having selected and
biased the BJT at a desired operating point, the effect of temperature must also be
taken into account. Temperature causes the device parameters such as the transistor
current gain (ac) and the transistor leakage current (ICEO) to change. Higher temperatures result in increased leakage currents in the device, thereby changing the operating condition set by the biasing network. The result is that the network design
must also provide a degree of temperature stability so that temperature changes result in minimum changes in the operating point. This maintenance of the operating
point can be specified by a stability factor, S, which indicates the degree of change
in operating point due to a temperature variation. A highly stable circuit is desirable,
and the stability of a few basic bias circuits will be compared.
For the BJT to be biased in its linear or active operating region the following must
be true:
1. The base–emitter junction must be forward-biased (p-region voltage more positive), with a resulting forward-bias voltage of about 0.6 to 0.7 V.
2. The base–collector junction must be reverse-biased (n-region more positive), with
the reverse-bias voltage being any value within the maximum limits of the device.
[Note that for forward bias the voltage across the p-n junction is p-positive, while for
reverse bias it is opposite (reverse) with n-positive. This emphasis on the initial letter should provide a means of helping memorize the necessary voltage polarity.]
Operation in the cutoff, saturation, and linear regions of the BJT characteristic are
provided as follows:
1. Linear-region operation:
Base–emitter junction forward biased
Base–collector junction reverse biased
4.2 Operating Point
145
2. Cutoff-region operation:
Base–emitter junction reverse biased
3. Saturation-region operation:
Base–emitter junction forward biased
Base–collector junction forward biased
4.3 FIXED-BIAS CIRCUIT
The fixed-bias circuit of Fig. 4.2 provides a relatively straightforward and simple introduction to transistor dc bias analysis. Even though the network employs an npn
transistor, the equations and calculations apply equally well to a pnp transistor configuration merely by changing all current directions and voltage polarities. The current directions of Fig. 4.2 are the actual current directions, and the voltages are defined by the standard double-subscript notation. For the dc analysis the network can
be isolated from the indicated ac levels by replacing the capacitors with an opencircuit equivalent. In addition, the dc supply VCC can be separated into two supplies
(for analysis purposes only) as shown in Fig. 4.3 to permit a separation of input and
output circuits. It also reduces the linkage between the two to the base current IB. The
separation is certainly valid, as we note in Fig. 4.3 that VCC is connected directly to
RB and RC just as in Fig. 4.2.
VCC
IC
RC
RB
C
ac
input
signal
IB
C1
B
C2
+
ac
output
signal
VCE
+
VBE
–
–
E
Figure 4.2 Fixed-bias circuit.
Figure 4.3 dc equivalent of
Fig. 4.2.
Forward Bias of Base–Emitter
Consider first the base–emitter circuit loop of Fig. 4.4. Writing Kirchhoff’s voltage
equation in the clockwise direction for the loop, we obtain
VCC IBRB VBE 0
Note the polarity of the voltage drop across RB as established by the indicated direction of IB. Solving the equation for the current IB will result in the following:
IB VCC VBE
RB
(4.4)
Figure 4.4 Base–emitter loop.
Equation (4.4) is certainly not a difficult one to remember if one simply keeps in
mind that the base current is the current through RB and by Ohm’s law that current
is the voltage across RB divided by the resistance RB. The voltage across RB is the applied voltage VCC at one end less the drop across the base-to-emitter junction (VBE).
146
Chapter 4
DC Biasing—BJTs
In addition, since the supply voltage VCC and the base–emitter voltage VBE are constants, the selection of a base resistor, RB, sets the level of base current for the operating point.
Collector–Emitter Loop
The collector–emitter section of the network appears in Fig. 4.5 with the indicated
direction of current IC and the resulting polarity across RC. The magnitude of the collector current is related directly to IB through
IC IB
(4.5)
It is interesting to note that since the base current is controlled by the level of RB
and IC is related to IB by a constant , the magnitude of IC is not a function of the
resistance RC. Change RC to any level and it will not affect the level of IB or IC as
long as we remain in the active region of the device. However, as we shall see, the
level of RC will determine the magnitude of VCE, which is an important parameter.
Applying Kirchhoff’s voltage law in the clockwise direction around the indicated
closed loop of Fig. 4.5 will result in the following:
Figure 4.5 Collector–emitter
loop.
VCE ICRC VCC 0
and
VCE VCC ICRC
(4.6)
which states in words that the voltage across the collector–emitter region of a transistor in the fixed-bias configuration is the supply voltage less the drop across RC.
As a brief review of single- and double-subscript notation recall that
VCE VC VE
(4.7)
where VCE is the voltage from collector to emitter and VC and VE are the voltages
from collector and emitter to ground respectively. But in this case, since VE 0 V,
we have
VCE VC
(4.8)
VBE VB VE
(4.9)
VBE VB
(4.10)
In addition, since
and VE 0 V, then
Keep in mind that voltage levels such as VCE are determined by placing the red
(positive) lead of the voltmeter at the collector terminal with the black (negative) lead
at the emitter terminal as shown in Fig. 4.6. VC is the voltage from collector to ground
and is measured as shown in the same figure. In this case the two readings are identical, but in the networks to follow the two can be quite different. Clearly understanding the difference between the two measurements can prove to be quite important in the troubleshooting of transistor networks.
Determine the following for the fixed-bias configuration of Fig. 4.7.
(a) IBQ and ICQ.
(b) VCEQ.
(c) VB and VC.
(d) VBC.
4.3 Fixed-Bias Circuit
Figure 4.6 Measuring VCE and
VC.
EXAMPLE 4.1
147
Figure 4.7 dc fixed-bias circuit for Example 4.1.
Solution
VCC VBE
12 V 0.7 V
47.08 A
RB
240 k
Eq. (4.5): ICQ IBQ (50)(47.08 A) 2.35 mA
(b) Eq. (4.6): VCEQ VCC ICRC
12 V (2.35 mA)(2.2 k )
6.83 V
(c) VB VBE 0.7 V
VC VCE 6.83 V
(d) Using double-subscript notation yields
(a) Eq. (4.4): IBQ VBC VB VC 0.7 V 6.83 V
6.13 V
with the negative sign revealing that the junction is reversed-biased, as it should be
for linear amplification.
Transistor Saturation
The term saturation is applied to any system where levels have reached their maximum values. A saturated sponge is one that cannot hold another drop of liquid. For
a transistor operating in the saturation region, the current is a maximum value for the
particular design. Change the design and the corresponding saturation level may rise
or drop. Of course, the highest saturation level is defined by the maximum collector
current as provided by the specification sheet.
Saturation conditions are normally avoided because the base–collector junction is
no longer reverse-biased and the output amplified signal will be distorted. An operating point in the saturation region is depicted in Fig. 4.8a. Note that it is in a region
where the characteristic curves join and the collector-to-emitter voltage is at or below VCEsat. In addition, the collector current is relatively high on the characteristics.
If we approximate the curves of Fig. 4.8a by those appearing in Fig. 4.8b, a quick,
direct method for determining the saturation level becomes apparent. In Fig. 4.8b, the
current is relatively high and the voltage VCE is assumed to be zero volts. Applying
Ohm’s law the resistance between collector and emitter terminals can be determined
as follows:
0V
V
0
RCE CE ICsat
IC
148
Chapter 4
DC Biasing—BJTs
IC
IC
I C sat –
0
Q-point
I C sat –
0
VCE
VCE sat
Q-point
VCE
(a)
(b)
Figure 4.8 Saturation regions: (a) actual; (b) approximate.
Applying the results to the network schematic would result in the configuration of
Fig. 4.9.
For the future, therefore, if there were an immediate need to know the approximate maximum collector current (saturation level) for a particular design, simply insert a short-circuit equivalent between collector and emitter of the transistor and calculate the resulting collector current. In short, set VCE 0 V. For the fixed-bias
configuration of Fig. 4.10, the short circuit has been applied, causing the voltage across
RC to be the applied voltage VCC. The resulting saturation current for the fixed-bias
configuration is
ICsat VCC
RC
Figure 4.9 Determining ICsat.
(4.11)
Figure 4.10 Determining ICsat
for the fixed-bias configuration.
Once ICsat is known, we have some idea of the maximum possible collector current
for the chosen design and the level to stay below if we expect linear amplification.
EXAMPLE 4.2
Determine the saturation level for the network of Fig. 4.7.
Solution
ICsat VCC
12 V
5.45 mA
RC
2.2 k
4.3 Fixed-Bias Circuit
149
The design of Example 4.1 resulted in ICQ 2.35 mA, which is far from the saturation level and about one-half the maximum value for the design.
Load-Line Analysis
The analysis thus far has been performed using a level of corresponding with the
resulting Q-point. We will now investigate how the network parameters define the
possible range of Q-points and how the actual Q-point is determined. The network of
Fig. 4.11a establishes an output equation that relates the variables IC and VCE in the
following manner:
VCE VCC ICRC
(4.12)
The output characteristics of the transistor also relate the same two variables IC and
VCE as shown in Fig. 4.11b.
In essence, therefore, we have a network equation and a set of characteristics that
employ the same variables. The common solution of the two occurs where the constraints established by each are satisfied simultaneously. In other words, this is similar to finding the solution of two simultaneous equations: one established by the network and the other by the device characteristics.
The device characteristics of IC versus VCE are provided in Fig. 4.11b. We must
now superimpose the straight line defined by Eq. (4.12) on the characteristics. The
most direct method of plotting Eq. (4.12) on the output characteristics is to use the
fact that a straight line is defined by two points. If we choose IC to be 0 mA, we are
specifying the horizontal axis as the line on which one point is located. By substituting IC 0 mA into Eq. (4.12), we find that
VCE VCC (0)RC
VCE VCC IC0 mA
and
(4.13)
defining one point for the straight line as shown in Fig. 4.12.
IC (mA)
50 µA
8
7
40 µA
6
30 µA
5
V CC
IC
+
RB
4
RC
3
+
2
20 µA
–
VCE
IB
–
10 µA
I B = 0 µA
1
0
5
10
15
ICEO
(a)
(b)
Figure 4.11 Load-line analysis: (a) the network; (b) the device characteristics.
150
Chapter 4
DC Biasing—BJTs
VCE (V)
IC
VCC
RC
Q-point
IB
VCE = 0 V
Q
Load line
0
VCC
IC = 0 mA
VCE
Figure 4.12 Fixed-bias
load line.
If we now choose VCE to be 0 V, which establishes the vertical axis as the line on
which the second point will be defined, we find that IC is determined by the following equation:
0 VCC ICRC
and
IC VCC
RC
VCE 0 V
(4.14)
as appearing on Fig. 4.12.
By joining the two points defined by Eqs. (4.13) and (4.14), the straight line established by Eq. (4.12) can be drawn. The resulting line on the graph of Fig. 4.12 is
called the load line since it is defined by the load resistor RC. By solving for the resulting level of IB, the actual Q-point can be established as shown in Fig. 4.12.
If the level of IB is changed by varying the value of RB the Q-point moves up or
down the load line as shown in Fig. 4.13. If VCC is held fixed and RC changed, the
load line will shift as shown in Fig. 4.14. If IB is held fixed, the Q-point will move
as shown in the same figure. If RC is fixed and VCC varied, the load line shifts as
shown in Fig. 4.15.
Figure 4.13 Movement of Q-point with increasing levels of IB.
Figure 4.14 Effect of increasing levels of RC on the load
line and Q-point.
4.3 Fixed-Bias Circuit
151
Figure 4.15 Effect of lower
values of VCC on the load line
and Q-point.
EXAMPLE 4.3
Given the load line of Fig. 4.16 and the defined Q-point, determine the required values of VCC, RC, and RB for a fixed-bias configuration.
I C (mA)
60 µA
12
50 µA
10
40 µA
8
30 µA
6
Q-point
20 µA
4
10 µA
2
0
I B = 0 µA
5
10
15
20
VCE
Figure 4.16 Example 4.3
Solution
From Fig. 4.16,
VCE VCC 20 V at IC 0 mA
and
and
152
Chapter 4
IC VCC
at VCE 0 V
RC
RC 20 V
VCC
2 k
10
mA
IC
IB VCC VBE
RB
RB VCC VBE
20 V 0.7 V
772 k
IB
25 A
DC Biasing—BJTs
4.4 EMITTER-STABILIZED BIAS CIRCUIT
The dc bias network of Fig. 4.17 contains an emitter resistor to improve the stability
level over that of the fixed-bias configuration. The improved stability will be demonstrated through a numerical example later in the section. The analysis will be performed by first examining the base–emitter loop and then using the results to investigate the collector–emitter loop.
Figure 4.17 BJT bias circuit
with emitter resistor.
Base–Emitter Loop
The base–emitter loop of the network of Fig. 4.17 can be redrawn as shown in Fig.
4.18. Writing Kirchhoff’s voltage law around the indicated loop in the clockwise direction will result in the following equation:
VCC IBRB VBE IERE 0
(4.15)
Recall from Chapter 3 that
IE ( 1)IB
(4.16)
Substituting for IE in Eq. (4.15) will result in
VCC IBRB VBE ( I)IBRE 0
Grouping terms will then provide the following:
IB(RB ( 1)RE) VCC VBE 0
Multiplying through by (1) we have
with
IB(RB ( 1)RE)VCC VBE 0
IB(RB ( 1)RE) VCC VBE
and solving for IB gives
IB VCC VBE
RB ( 1)RE
(4.17)
Note that the only difference between this equation for IB and that obtained for the
fixed-bias configuration is the term ( 1)RE.
There is an interesting result that can be derived from Eq. (4.17) if the equation
is used to sketch a series network that would result in the same equation. Such is
4.4 Emitter-Stabilized Bias Circuit
Figure 4.18 Base–emitter loop.
153
Figure 4.20 Reflected impedance
level of RE.
Figure 4.19 Network derived
from Eq. (4.17).
the case for the network of Fig. 4.19. Solving for the current IB will result in the same
equation obtained above. Note that aside from the base-to-emitter voltage VBE, the
resistor RE is reflected back to the input base circuit by a factor ( 1). In other
words, the emitter resistor, which is part of the collector–emitter loop, “appears as”
( 1)RE in the base–emitter loop. Since is typically 50 or more, the emitter resistor appears to be a great deal larger in the base circuit. In general, therefore, for
the configuration of Fig. 4.20,
Ri ( 1)RE
(4.18)
Equation (4.18) is one that will prove useful in the analysis to follow. In fact, it
provides a fairly easy way to remember Eq. (4.17). Using Ohm’s law, we know that
the current through a system is the voltage divided by the resistance of the circuit.
For the base–emitter circuit the net voltage is VCC VBE. The resistance levels are
RB plus RE reflected by ( 1). The result is Eq. (4.17).
Collector–Emitter Loop
The collector–emitter loop is redrawn in Fig. 4.21. Writing Kirchhoff’s voltage law
for the indicated loop in the clockwise direction will result in
IERE VCE ICRC VCC 0
Substituting IE IC and grouping terms gives
VCE VCC IC (RC RE) 0
VCE VCC IC (RC RE)
and
(4.19)
The single-subscript voltage VE is the voltage from emitter to ground and is determined by
VE IERE
Figure 4.21 Collector–emitter
loop.
(4.20)
while the voltage from collector to ground can be determined from
VCE VC VE
and
or
VC VCE VE
(4.21)
VC VCC ICRC
(4.22)
The voltage at the base with respect to ground can be determined from
or
154
Chapter 4
DC Biasing—BJTs
VB VCC IBRB
(4.23)
VB VBE VE
(4.24)
For the emitter bias network of Fig. 4.22, determine:
(a) IB.
(b) IC.
(c) VCE.
(d) VC.
(e) VE.
(f) VB.
(g) VBC.
EXAMPLE 4.4
Figure 4.22 Emitter-stabilized
bias circuit for Example 4.4.
Solution
20 V 0.7 V
VCC VBE
430 k (51)(1 k )
RB ( 1)RE
19.3 V
40.1 A
481 k
(a) Eq. (4.17): IB (b) IC IB
(50)(40.1 A)
2.01 mA
(c) Eq. (4.19): VCE VCC IC (RC RE)
20 V (2.01 mA)(2 k 1 k ) 20 V 6.03 V
13.97 V
(d) VC VCC ICRC
20 V (2.01 mA)(2 k ) 20 V 4.02 V
15.98 V
(e) VE VC VCE
15.98 V 13.97 V
2.01 V
or VE IERE ICRE
(2.01 mA)(1 k )
2.01 V
(f) VB VBE VE
0.7 V 2.01 V
2.71 V
(g) VBC VB VC
2.71 V 15.98 V
13.27 V (reverse-biased as required)
4.4 Emitter-Stabilized Bias Circuit
155
Improved Bias Stability
The addition of the emitter resistor to the dc bias of the BJT provides improved stability, that is, the dc bias currents and voltages remain closer to where they were set
by the circuit when outside conditions, such as temperature, and transistor beta,
change. While a mathematical analysis is provided in Section 4.12, some comparison
of the improvement can be obtained as demonstrated by Example 4.5.
EXAMPLE 4.5
Prepare a table and compare the bias voltage and currents of the circuits of Figs. 4.7
and Fig. 4.22 for the given value of 50 and for a new value of 100. Compare the changes in IC and VCE for the same increase in .
Solution
Using the results calculated in Example 4.1 and then repeating for a value of 100
yields the following:
IB (A)
IC (mA)
VCE (V)
50
100
47.08
47.08
2.35
4.71
6.83
1.64
The BJT collector current is seen to change by 100% due to the 100% change in the
value of . IB is the same and VCE decreased by 76%.
Using the results calculated in Example 4.4 and then repeating for a value of
100, we have the following:
IB (A)
IC (mA)
VCE (V)
50
100
40.1
36.3
2.01
3.63
13.97
9.11
Now the BJT collector current increases by about 81% due to the 100% increase in .
Notice that IB decreased, helping maintain the value of IC —or at least reducing the
overall change in IC due to the change in . The change in VCE has dropped to about
35%. The network of Fig. 4.22 is therefore more stable than that of Fig. 4.7 for the
same change in .
Saturation Level
The collector saturation level or maximum collector current for an emitter-bias design can be determined using the same approach applied to the fixed-bias configuration: Apply a short circuit between the collector–emitter terminals as shown in Fig.
4.23 and calculate the resulting collector current. For Fig. 4.23:
ICsat VCC
RC RE
(4.25)
Figure 4.23 Determining ICsat for
the emitter-stabilized bias circuit.
The addition of the emitter resistor reduces the collector saturation level below that
obtained with a fixed-bias configuration using the same collector resistor.
156
Chapter 4
DC Biasing—BJTs
EXAMPLE 4.6
Determine the saturation current for the network of Example 4.4.
Solution
ICsat VCC
RC RE
20 V
2 k 1 k
20 V
3k
6.67 mA
which is about twice the level of ICQ for Example 4.4.
Load-Line Analysis
The load-line analysis of the emitter-bias network is only slightly different from that
encountered for the fixed-bias configuration. The level of IB as determined by Eq.
(4.17) defines the level of IB on the characteristics of Fig. 4.24 (denoted IBQ).
Figure 4.24 Load line for the
emitter-bias configuration.
The collector–emitter loop equation that defines the load line is the following:
VCE VCC IC (RC RE)
Choosing IC 0 mA gives
VCE VCCIC0 mA
(4.26)
as obtained for the fixed-bias configuration. Choosing VCE 0 V gives
IC VCC
RC RE
VCE0 V
(4.27)
as shown in Fig. 4.24. Different levels of IBQ will, of course, move the Q-point up or
down the load line.
4.5 VOLTAGE-DIVIDER BIAS
In the previous bias configurations the bias current ICQ and voltage VCEQ were a function of the current gain () of the transistor. However, since is temperature sensitive, especially for silicon transistors, and the actual value of beta is usually not well
defined, it would be desirable to develop a bias circuit that is less dependent, or in
4.5 Voltage-Divider Bias
157
Figure 4.26 Defining the Q-point for the voltage-divider
bias configuration.
Figure 4.25 Voltage-divider bias configuration.
fact, independent of the transistor beta. The voltage-divider bias configuration of Fig.
4.25 is such a network. If analyzed on an exact basis the sensitivity to changes in beta
is quite small. If the circuit parameters are properly chosen, the resulting levels of ICQ
and VCEQ can be almost totally independent of beta. Recall from previous discussions
that a Q-point is defined by a fixed level of ICQ and VCEQ as shown in Fig. 4.26. The
level of IBQ will change with the change in beta, but the operating point on the characteristics defined by ICQ and VCEQ can remain fixed if the proper circuit parameters
are employed.
As noted above, there are two methods that can be applied to analyze the voltagedivider configuration. The reason for the choice of names for this configuration will
become obvious in the analysis to follow. The first to be demonstrated is the exact
method that can be applied to any voltage-divider configuration. The second is referred to as the approximate method and can be applied only if specific conditions
are satisfied. The approximate approach permits a more direct analysis with a savings
in time and energy. It is also particularly helpful in the design mode to be described
in a later section. All in all, the approximate approach can be applied to the majority
of situations and therefore should be examined with the same interest as the exact
method.
Exact Analysis
The input side of the network of Fig. 4.25 can be redrawn as shown in Fig. 4.27 for
the dc analysis. The Thévenin equivalent network for the network to the left of the
base terminal can then be found in the following manner:
B
R1
VCC
R2
RE
Thévenin
158
Chapter 4
DC Biasing—BJTs
Figure 4.27 Redrawing the
input side of the network of
Fig. 4.25.
RTh: The voltage source is replaced by a short-circuit equivalent as shown in
Fig. 4.28.
RTh R1R2
R2
(4.28)
ETh: The voltage source VCC is returned to the network and the open-circuit
Thévenin voltage of Fig. 4.29 determined as follows:
Applying the voltage-divider rule:
ETh VR2 R1
R2VCC
R1 R2
R Th
Figure 4.28 Determining RTh.
(4.29)
The Thévenin network is then redrawn as shown in Fig. 4.30, and IBQ can be determined by first applying Kirchhoff’s voltage law in the clockwise direction for the
loop indicated:
VCC
+
+
R1
R2
VR E Th
2
–
–
ETh IBRTh VBE IERE 0
Substituting IE ( 1)IB and solving for IB yields
IB Figure 4.29 Determining ETh.
ETh VBE
RTh ( 1)RE
(4.30)
Although Eq. (4.30) initially appears different from those developed earlier, note
that the numerator is again a difference of two voltage levels and the denominator is
the base resistance plus the emitter resistor reflected by ( 1)—certainly very similar to Eq. (4.17).
Once IB is known, the remaining quantities of the network can be found in the
same manner as developed for the emitter-bias configuration. That is,
VCE VCC IC (RC RE)
RTh
B
+
IB
VBE
–
ETh
E
RE
(4.31)
which is exactly the same as Eq. (4.19). The remaining equations for VE, VC, and VB
are also the same as obtained for the emitter-bias configuration.
Figure 4.30 Inserting the
Thévenin equivalent circuit.
Determine the dc bias voltage VCE and the current IC for the voltage-divider configuration of Fig. 4.31.
EXAMPLE 4.7
Figure 4.31 Beta-stabilized
circuit for Example 4.7.
4.5 Voltage-Divider Bias
159
Solution
Eq. (4.28):
RTh R1R2
Eq. (4.29):
ETh (39 k )(3.9 k )
3.55 k
39 k 3.9 k
R2VCC
R1 R2
(3.9 k )(22 V)
2V
39 k 3.9 k
ETh VBE
IB RTh ( 1)RE
Eq. (4.30):
3.55 k
2 V 0.7 V
1.3 V
(141)(1.5 k )
3.55 k 211.5 k
6.05 A
IC IB
(140)(6.05 A)
0.85 mA
Eq. (4.31):
VCE VCC IC (RC RE)
22 V (0.85 mA)(10 k
1.5 k )
22 V 9.78 V
12.22 V
Approximate Analysis
The input section of the voltage-divider configuration can be represented by the network of Fig. 4.32. The resistance Ri is the equivalent resistance between base and
ground for the transistor with an emitter resistor RE. Recall from Section 4.4 [Eq.
(4.18)] that the reflected resistance between base and emitter is defined by Ri ( 1)RE. If Ri is much larger than the resistance R2, the current IB will be much
smaller than I2 (current always seeks the path of least resistance) and I2 will be approximately equal to I1. If we accept the approximation that IB is essentially zero amperes compared to I1 or I2, then I1 I2 and R1 and R2 can be considered series ele-
Figure 4.32 Partial-bias circuit
for calculating the approximate
base voltage VB.
160
Chapter 4
DC Biasing—BJTs
ments. The voltage across R2, which is actually the base voltage, can be determined
using the voltage-divider rule (hence the name for the configuration). That is,
VB R2VCC
R1 R2
(4.32)
Since Ri ( 1)RE RE the condition that will define whether the approximate approach can be applied will be the following:
RE
10R2
(4.33)
In other words, if times the value of RE is at least 10 times the value of R2, the approximate approach can be applied with a high degree of accuracy.
Once VB is determined, the level of VE can be calculated from
VE VB VBE
(4.34)
and the emitter current can be determined from
VE
RE
(4.35)
ICQ IE
(4.36)
IE and
The collector-to-emitter voltage is determined by
VCE VCC ICRC IERE
but since IE IC,
VCEQ VCC IC (RC RE)
(4.37)
Note in the sequence of calculations from Eq. (4.33) through Eq. (4.37) that does not appear and IB was not calculated. The Q-point (as determined by ICQ and
VCEQ) is therefore independent of the value of .
Repeat the analysis of Fig. 4.31 using the approximate technique, and compare solutions for ICQ and VCEQ.
EXAMPLE 4.8
Solution
Testing:
RE
(140)(1.5 k )
210 k
Eq. (4.32):
10R2
10(3.9 k )
39 k
VB (satisfied)
R2VCC
R1 R2
(3.9 k )(22 V)
39 k 3.9 k
2V
4.5 Voltage-Divider Bias
161
Note that the level of VB is the same as ETh determined in Example 4.7. Essentially, therefore, the primary difference between the exact and approximate techniques
is the effect of RTh in the exact analysis that separates ETh and VB.
Eq. (4.34):
VE VB VBE
2 V 0.7 V
1.3 V
ICQ IE VE
1.3 V
0.867 mA
RE
1.5 k
compared to 0.85 mA with the exact analysis. Finally,
VCEQ VCC IC(RC RE)
22 V (0.867 mA)(10 kV 1.5 k )
22 V 9.97 V
12.03 V
versus 12.22 V obtained in Example 4.7.
The results for ICQ and VCEQ are certainly close, and considering the actual variation in parameter values one can certainly be considered as accurate as the other.
The larger the level of Ri compared to R2, the closer the approximate to the exact solution. Example 4.10 will compare solutions at a level well below the condition established by Eq. (4.33).
EXAMPLE 4.9
Repeat the exact analysis of Example 4.7 if is reduced to 70, and compare solutions for ICQ and VCEQ.
Solution
This example is not a comparison of exact versus approximate methods but a testing
of how much the Q-point will move if the level of is cut in half. RTh and ETh are
the same:
RTh 3.55 k ,
IB ETh 2 V
ETh VBE
RTh ( 1)RE
2 V 0.7 V
1.3 V
3.55 k (71)(1.5 k )
3.55 k 106.5 k
11.81 A
ICQ IB
(70)(11.81 A)
0.83 mA
VCEQ VCC IC(RC RE)
22 V (0.83 mA)(10 k
12.46 V
162
Chapter 4
DC Biasing—BJTs
1.5 k )
Tabulating the results, we have:
ICQ (mA)
VCEQ (V)
140
70
0.85
0.83
12.22
12.46
The results clearly show the relative insensitivity of the circuit to the change in .
Even though is drastically cut in half, from 140 to 70, the levels of ICQ and VCEQ
are essentially the same.
Determine the levels of ICQ and VCEQ for the voltage-divider configuration of Fig. 4.33
using the exact and approximate techniques and compare solutions. In this case, the
conditions of Eq. (4.33) will not be satisfied but the results will reveal the difference
in solution if the criterion of Eq. (4.33) is ignored.
EXAMPLE 4.10
Figure 4.33 Voltage-divider
configuration for Example 4.10.
Solution
Exact Analysis
Eq. (4.33):
RE
(50)(1.2 k )
60 k
10R2
10(22 k )
220 k
RTh R1R2 82 k 22 k
ETh IB (not satisfied)
17.35 k
R2VCC
22 k (18 V)
3.81 V
R1 R2
82 k 22 k
3.81 V 0.7 V
3.11 V
ETh VBE
17.35
k
(51)(1.2
k
)
RTh ( 1)RE
78.55 k
39.6 A
ICQ IB (50)(39.6 A) 1.98 mA
VCEQ VCC IC(RC RE)
18 V (1.98 mA)(5.6 k
1.2 k )
4.54 V
4.5 Voltage-Divider Bias
163
Approximate Analysis
VB ETh 3.81 V
VE VB VBE 3.81 V 0.7 V 3.11 V
ICQ IE VE
3.11 V
2.59 mA
RE
1.2 k
VCEQ VCC IC (RC RE)
18 V (2.59 mA)(5.6 k
1.2 k )
3.88 V
Tabulating the results, we have:
Exact
Approximate
ICQ (mA)
VCEQ (V)
1.98
2.59
4.54
3.88
The results reveal the difference between exact and approximate solutions. ICQ is about
30% greater with the approximate solution, while VCEQ is about 10% less. The results
are notably different in magnitude, but even though RE is only about three times
larger than R2, the results are still relatively close to each other. For the future, however, our analysis will be dictated by Eq. (4.33) to ensure a close similarity between
exact and approximate solutions.
Transistor Saturation
The output collector–emitter circuit for the voltage-divider configuration has the same
appearance as the emitter-biased circuit analyzed in Section 4.4. The resulting equation for the saturation current (when VCE is set to zero volts on the schematic) is therefore the same as obtained for the emitter-biased configuration. That is,
VCC
RC RE
ICsat ICmax (4.38)
Load-Line Analysis
The similarities with the output circuit of the emitter-biased configuration result in
the same intersections for the load line of the voltage-divider configuration. The load
line will therefore have the same appearance as that of Fig. 4.24, with
IC VCC
RC RE VCE0 V
VCE VCCIC0 mA
and
(4.39)
(4.40)
The level of IB is of course determined by a different equation for the voltage-divider
bias and the emitter-bias configurations.
164
Chapter 4
DC Biasing—BJTs
4.6 DC BIAS WITH VOLTAGE FEEDBACK
An improved level of stability can also be obtained by introducing a feedback path
from collector to base as shown in Fig. 4.34. Although the Q-point is not totally independent of beta (even under approximate conditions), the sensitivity to changes in
beta or temperature variations is normally less than encountered for the fixed-bias or
emitter-biased configurations. The analysis will again be performed by first analyzing the base–emitter loop with the results applied to the collector–emitter loop.
Base–Emitter Loop
Figure 4.35 shows the base–emitter loop for the voltage feedback configuration. Writing Kirchhoff’s voltage law around the indicated loop in the clockwise direction will
result in
VCC I CRC IBRB VBE IERE 0
VCC
+
RC
–
I C'
vo
RB
IC
IB
C2
+
vi
RB
VCC
RC
I C'
IC
IB
+
VCE
–
C1
+–
VBE
IE
–
+
IE
RE
RE
–
Figure 4.35 Base–emitter loop for the
network of Fig. 4.34.
Figure 4.34 dc bias circuit with voltage feedback.
It is important to note that the current through RC is not IC but I C (where I C IC IB). However, the level of IC and IC far exceeds the usual level of IB and the approximation I C IC is normally employed. Substituting I C IC IB and IE IC
will result in
VCC IBRC IBRB VBE IBRE 0
Gathering terms, we have
VCC VBE IB(RC RE) IBRB 0
and solving for IB yields
IB VCC VBE
RB (RC RE)
(4.41)
The result is quite interesting in that the format is very similar to equations for IB
obtained for earlier configurations. The numerator is again the difference of available
voltage levels, while the denominator is the base resistance plus the collector and emitter resistors reflected by beta. In general, therefore, the feedback path results in a reflection of the resistance RC back to the input circuit, much like the reflection of RE.
In general, the equation for IB has had the following format:
IB V
RB R
4.6 DC Bias with Voltage Feedback
165
with the absence of R for the fixed-bias configuration, R RE for the emitter-bias
setup (with ( 1) ), and R RC RE for the collector-feedback arrangement.
The voltage V is the difference between two voltage levels.
Since IC IB,
V
RB R
ICQ In general, the larger R is compared to RB, the less the sensitivity of ICQ to variations in beta. Obviously, if R RB and RB R R , then
ICQ I'C
and ICQ is independent of the value of beta. Since R is typically larger for the voltagefeedback configuration than for the emitter-bias configuration, the sensitivity to variations in beta is less. Of course, R is zero ohms for the fixed-bias configuration and
is therefore quite sensitive to variations in beta.
+
RC
Collector–Emitter Loop
–
IC
+
The collector–emitter loop for the network of Fig. 4.34 is provided in Fig. 4.36. Applying Kirchhoff’s voltage law around the indicated loop in the clockwise direction
will result in
VCC
IERE VCE I CRC VCC 0
VCE
IE
V
V
V
RB R
R
R
–
+
Since I C IC and IE IC, we have
IC (RC RE) VCE VCC 0
RE
–
VCE VCC IC (RC RE)
and
(4.42)
which is exactly as obtained for the emitter-bias and voltage-divider bias configurations.
Figure 4.36 Collector–emitter
loop for the network of Fig. 4.34.
EXAMPLE 4.11
Determine the quiescent levels of ICQ and VCEQ for the network of Fig. 4.37.
Solution
Eq. (4.41):
IB 10 V
4.7 kΩ
250 kΩ
β = 90
vi
10 V 0.7 V
(90)(4.7 k 1.2 k )
250 k
250 k
9.3 V
531 k
9.3 V
781 k
11.91 A
vo
10 µF
VCC VBE
RB (RC RE)
ICQ IB (90)(11.91 A)
1.07 mA
10 µF
VCEQ VCC IC (RC RE)
1.2 kΩ
10 V (1.07 mA)(4.7 k
10 V 6.31 V
3.69 V
Figure 4.37 Network for Example 4.11.
166
Chapter 4
DC Biasing—BJTs
1.2 k )
Repeat Example 4.11 using a beta of 135 (50% more than Example 4.11).
EXAMPLE 4.12
Solution
It is important to note in the solution for IB in Example 4.11 that the second term in
the denominator of the equation is larger than the first. Recall in a recent discussion
that the larger this second term is compared to the first, the less the sensitivity to
changes in beta. In this example the level of beta is increased by 50%, which will increase the magnitude of this second term even more compared to the first. It is more
important to note in these examples, however, that once the second term is relatively
large compared to the first, the sensitivity to changes in beta is significantly less.
Solving for IB gives
IB VCC VBE
RB (RC RE)
10 V 0.7 V
(135)(4.7 k 1.2 k )
250 k
250 k
9.3 V
796.5 k
9.3 V
1046.5 k
8.89 A
ICQ IB
and
(135)(8.89 A)
1.2 mA
and
VCEQ VCC IC (RC RE)
10 V (1.2 mA)(4.7 k
1.2 k )
10 V 7.08 V
2.92 V
Even though the level of increased 50%, the level of ICQ only increased 12.1%
while the level of VCEQ decreased about 20.9%. If the network were a fixed-bias design, a 50% increase in would have resulted in a 50% increase in ICQ and a dramatic change in the location of the Q-point.
Determine the dc level of IB and VC for the network of Fig. 4.38.
EXAMPLE 4.13
18 V
91 kΩ
110 kΩ
3.3 kΩ
10 µF
vo
R1
R2
10 µF
10 µF
β = 75
vi
510 Ω
50 µF
Figure 4.38 Network for
Example 4.13.
4.6 DC Bias with Voltage Feedback
167
Solution
In this case, the base resistance for the dc analysis is composed of two resistors with
a capacitor connected from their junction to ground. For the dc mode, the capacitor
assumes the open-circuit equivalence and RB R1 R2.
Solving for IB gives
IB VCC VBE
RB (RC RE)
(91 k
18 V 0.7 V
110 k ) (75)(3.3 k
201 k
17.3 V
285.75 k
0.51 k )
17.3 V
486.75 k
35.5 A
IC IB
(75)(35.5 A)
2.66 mA
VC VCC I CRC VCC ICRC
18 V (2.66 mA)(3.3 k )
18 V 8.78 V
9.22 V
Saturation Conditions
Using the approximation I C IC , the equation for the saturation current is the same
as obtained for the voltage-divider and emitter-bias configurations. That is,
ICsat ICmax VCC
RC RE
(4.43)
Load-Line Analysis
Continuing with the approximation I C IC will result in the same load line defined
for the voltage-divider and emitter-biased configurations. The level of IBQ will be defined by the chosen bias configuration.
4.7 MISCELLANEOUS BIAS
CONFIGURATIONS
There are a number of BJT bias configurations that do not match the basic mold of
those analyzed in the previous sections. In fact, there are variations in design that
would require many more pages than is possible in a book of this type. However, the
primary purpose here is to emphasize those characteristics of the device that permit
a dc analysis of the configuration and to establish a general procedure toward the desired solution. For each configuration discussed thus far, the first step has been the
derivation of an expression for the base current. Once the base current is known, the
collector current and voltage levels of the output circuit can be determined quite di168
Chapter 4
DC Biasing—BJTs
rectly. This is not to imply that all solutions will take this path, but it does suggest a
possible route to follow if a new configuration is encountered.
The first example is simply one where the emitter resistor has been dropped from
the voltage-feedback configuration of Fig. 4.34. The analysis is quite similar but does
require dropping RE from the applied equation.
EXAMPLE 4.14
For the network of Fig. 4.39:
(a) Determine ICQ and VCEQ.
(b) Find VB, VC, VE, and VBC.
Figure 4.39 Collector feedback
with RE 0 .
Solution
(a) The absence of RE reduces the reflection of resistive levels to simply that of RC
and the equation for IB reduces to
IB VCC VBE
RB RC
20 V 0.7 V
19.3 V
680 k (120)(4.7 k )
1.244 M
15.51 A
ICQ IB (120)(15.51 A)
1.86 mA
VCEQ VCC ICRC
20 V (1.86 mA)(4.7 k )
11.26 V
VB VBE 0.7 V
VC VCE 11.26 V
VE 0 V
VBC VB VC 0.7 V 11.26 V
10.56 V
In the next example, the applied voltage is connected to the emitter leg and RC is
connected directly to ground. Initially, it appears somewhat unorthodox and quite different from those encountered thus far, but one application of Kirchhoff’s voltage law
to the base circuit will result in the desired base current.
4.7 Miscellaneous Bias Configurations
169
EXAMPLE 4.15
Determine VC and VB for the network of Fig. 4.40.
Figure 4.40 Example 4.15
Solution
Applying Kirchhoff’s voltage law in the clockwise direction for the base–emitter loop
will result in
IBRB VBE VEE 0
IB and
VEE VBE
RB
Substitution yields
IB 9 V 0.7 V
100 k
8.3 V
100 k
83 A
IC IB
(45)(83 A)
3.735 mA
VC ICRC
(3.735 mA)(1.2 k )
4.48 V
VB IBRB
(83 A)(100 k )
8.3 V
The next example employs a network referred to as an emitter-follower configuration. When the same network is analyzed on an ac basis, we will find that the output and input signals are in phase (one following the other) and the output voltage is
slightly less than the applied signal. For the dc analysis the collector is grounded and
the applied voltage is in the emitter leg.
170
Chapter 4
DC Biasing—BJTs
EXAMPLE 4.16
Determine VCEQ and IE for the network of Fig. 4.41.
Figure 4.41 Common-collector (emitter-follower) configuration.
Solution
Applying Kirchhoff’s voltage law to the input circuit will result in
IBRB VBE IERE VEE 0
but
IE ( 1)IB
and
VEE VBE ( 1)IBRE IBRB 0
VEE VBE
IB RB ( 1)RE
with
Substituting values yields
IB 20 V 0.7 V
240 k (91)(2 k )
240 k
19.3 V
182 k
19.3 V
422 k
45.73 A
IC IB
(90)(45.73 A)
4.12 mA
Applying Kirchhoff’s voltage law to the output circuit, we have
VEE IERE VCE 0
but
and
IE ( 1)IB
VCEQ VEE ( 1)IBRE
20 V (91)(45.73 A)(2 k )
11.68 V
IE 4.16 mA
4.7 Miscellaneous Bias Configurations
171
All of the examples thus far have employed a common-emitter or commoncollector configuration. In the next example we investigate the common-base configuration. In this situation the input circuit will be employed to determine IE rather than
IB. The collector current is then available to perform an analysis of the output circuit.
EXAMPLE 4.17
Determine the voltage VCB and the current IB for the common-base configuration of
Fig. 4.42.
Figure 4.42 Common-base configuration.
Solution
Applying Kirchhoff’s voltage law to the input circuit yields
VEE IERE VBE 0
IE and
VEE VBE
RE
Substituting values, we obtain
IE 4 V 0.7 V
2.75 mA
1.2 k
Applying Kirchhoff’s voltage law to the output circuit gives
VCB ICRC VCC 0
and
VCB VCC ICRC with IC IE
10 V (2.75 mA)(2.4 k )
3.4 V
IB IC
2.75 mA
60
45.8 A
Example 4.18 employs a split supply and will require the application of Thévenin’s
theorem to determine the desired unknowns.
172
Chapter 4
DC Biasing—BJTs
EXAMPLE 4.18
Determine VC and VB for the network of Fig. 4.43.
VCC = + 20 V
RC
R1
2.7 kΩ
8.2 kΩ
C2
C
vo
10 µF
C1
B
vi
β = 120
10 µF
E
R2
2.2 kΩ
RE
1.8 kΩ
Figure 4.43 Example 4.18
VEE = – 20 V
Solution
The Thévenin resistance and voltage are determined for the network to the left of the
base terminal as shown in Figs. 4.44 and 4.45.
8.2 kΩ
R1
I
B
R1
+
R2
8.2 kΩ
R2
VCC
2.2 kΩ
VEE
RTh
Figure 4.44 Determining RTh.
2.2 kΩ
–
–
20 V
20 V
+
+
B
ETh
–
Figure 4.45 Determining ETh.
RTh:
RTh 8.2 k 2.2 k
1.73 k
ETh:
I
VCC VEE
20 V 20 V
R1 R2
8.2 k 2.2 k
40 V
10.4 k
3.85 mA
ETh IR2 VEE
(3.85 mA)(2.2 k ) 20 V
11.53 V
The network can then be redrawn as shown in Fig. 4.46, where the application of
Kirchhoff’s voltage law will result in
ETh IBRTh VBE IERE VEE 0
4.7 Miscellaneous Bias Configurations
173
+
R Th
–
VB
IB
E Th
β = 120
+
1.73 kΩ
VBE
11.53 V
–
RE
E
+
1.8 kΩ
–
VEE = –20 V
Figure 4.46 Substituting the
Thévenin equivalent circuit.
Substituting IE ( 1)IB gives
VEE ETh VBE ( 1)IBRE IBRTh 0
and
IB VEE ETh VBE
RTh ( 1)RE
20 V 11.53 V 0.7 V
1.73 k (121)(1.8 k )
7.77 V
219.53 k
35.39 A
IC IB
(120)(35.39 A)
4.25 mA
VC VCC ICRC
20 V (4.25 mA)(2.7 k )
8.53 V
VB ETh IBRTh
(11.53 V) (35.39 A)(1.73 k )
11.59 V
4.8 DESIGN OPERATIONS
Discussions thus far have focused on the analysis of existing networks. All the elements are in place and it is simply a matter of solving for the current and voltage levels of the configuration. The design process is one where a current and/or voltage may
be specified and the elements required to establish the designated levels must be determined. This synthesis process requires a clear understanding of the characteristics
of the device, the basic equations for the network, and a firm understanding of the
basic laws of circuit analysis, such as Ohm’s law, Kirchhoff’s voltage law, and so on.
In most situations the thinking process is challenged to a higher degree in the design
process than in the analysis sequence. The path toward a solution is less defined and
in fact may require a number of basic assumptions that do not have to be made when
simply analyzing a network.
174
Chapter 4
DC Biasing—BJTs
The design sequence is obviously sensitive to the components that are already
specified and the elements to be determined. If the transistor and supplies are specified, the design process will simply determine the required resistors for a particular
design. Once the theoretical values of the resistors are determined, the nearest standard commercial values are normally chosen and any variations due to not using the
exact resistance values are accepted as part of the design. This is certainly a valid approximation considering the tolerances normally associated with resistive elements
and the transistor parameters.
If resistive values are to be determined, one of the most powerful equations is
simply Ohm’s law in the following form:
Runk VR
IR
(4.44)
In a particular design the voltage across a resistor can often be determined from specified levels. If additional specifications define the current level, Eq. (4.44) can then
be used to calculate the required resistance level. The first few examples will demonstrate how particular elements can be determined from specified levels. A complete
design procedure will then be introduced for two popular configurations.
Given the device characteristics of Fig. 4.47a, determine VCC, RB, and RC for the fixedbias configuration of Fig. 4.47b.
EXAMPLE 4.19
Figure 4.47 Example 4.19
Solution
From the load line
and
with
VCC 20 V
IC VCC
RC
RC VCC
20 V
2.5 k
IC
8 mA
VCE 0 V
IB VCC VBE
RB
RB VCC VBE
IB
20 V 0.7 V
19.3 V
40 A
40 A
482.5 k
4.8 Design Operations
175
Standard resistor values:
RC 2.4 k
RB 470 k
Using standard resistor values gives
IB 41.1 A
which is well within 5% of the value specified.
EXAMPLE 4.20
Given that ICQ 2 mA and VCEQ 10 V, determine R1 and RC for the network of
Fig. 4.48.
Figure 4.48 Example 4.20
Solution
VE IERE ICRE
(2 mA)(1.2 k ) 2.4 V
VB VBE VE 0.7 V 2.4 V 3.1 V
VB and
R2VCC
3.1 V
R1 R2
(18 k )(18 V)
3.1 V
R1 18 k
324 k
3.1R1 55.8 k
3.1R1 268.2 k
R1 Eq. (4.44):
RC VRC
IC
268.2 k
3.1
86.52 k
VCC VC
IC
with
VC VCE VE 10 V 2.4 V 12.4 V
and
RC 18 V 12.4V
2mA
2.8 k
The nearest standard commercial values to R1 are 82 and 91 k . However, using
the series combination of standard values of 82 k and 4.7 k 86.7 k would result in a value very close to the design level.
176
Chapter 4
DC Biasing—BJTs
The emitter-bias configuration of Fig. 4.49 has the following specifications: ICQ 1
2 ICsat, ICsat 8 mA, VC 18 V, and 110. Determine RC, RE, and RB.
EXAMPLE 4.21
Figure 4.49 Example 4.21
Solution
ICQ 12 ICsat 4 mA
RC ICsat and
VRC
ICQ
VCC VC
ICQ
28 V 18 V
= 2.5 k
4 mA
VCC
RC RE
RC RE VCC
28 V
3.5 k
ICsat
8 mA
RE 3.5 k
3.5 k
RC
2.5 k
1 k
IBQ IBQ ICQ
4 mA
36.36 A
110
VCC VBE
RB ( 1)RE
and
RB ( 1)RE with
RB VCC VBE
IBQ
VCC VBE
( 1)RE
IBQ
28 V 0.7 V
(111)(1 k )
36.36 A
27.3 V
111 k
36.36 A
639.8 k
4.8 Design Operations
177
For standard values:
RC 2.4 k
RE 1 k
RB 620 k
The discussion to follow will introduce one technique for designing an entire circuit to operate at a specified bias point. Often the manufacturer’s specification (spec)
sheets provide information on a suggested operating point (or operating region) for a
particular transistor. In addition, other system components connected to the given amplifier stage may also define the current swing, voltage swing, value of common supply voltage, and so on, for the design.
In actual practice, many other factors may have to be considered that may affect
the selection of the desired operating point. For the moment we shall concentrate,
however, on determining the component values to obtain a specified operating point.
The discussion will be limited to the emitter-bias and voltage-divider bias configurations, although the same procedure can be applied to a variety of other transistor circuits.
Design of a Bias Circuit with an Emitter
Feedback Resistor
Consider first the design of the dc bias components of an amplifier circuit having
emitter-resistor bias stabilization as shown in Fig. 4.50. The supply voltage and operating point were selected from the manufacturer’s information on the transistor used
in the amplifier.
Figure 4.50 Emitter-stabilized
bias circuit for design consideration.
The selection of collector and emitter resistors cannot proceed directly from
the information just specified. The equation that relates the voltages around the
collector–emitter loop has two unknown quantities present—the resistors RC and RE.
At this point some engineering judgment must be made, such as the level of the emitter voltage compared to the applied supply voltage. Recall that the need for including a resistor from emitter to ground was to provide a means of dc bias stabilization
so that the change of collector current due to leakage currents in the transistor and
the transistor beta would not cause a large shift in the operating point. The emitter
resistor cannot be unreasonably large because the voltage across it limits the range of
voltage swing of the voltage from collector to emitter (to be noted when the ac re178
Chapter 4
DC Biasing—BJTs
sponse is discussed). The examples examined in this chapter reveal that the voltage
from emitter to ground is typically around one-fourth to one-tenth of the supply voltage. Selecting the conservative case of one-tenth will permit calculating the emitter
resistor RE and the resistor RC in a manner similar to the examples just completed. In
the next example we perform a complete design of the network of Fig. 4.49 using the
criteria just introduced for the emitter voltage.
Determine the resistor values for the network of Fig. 4.50 for the indicated operating
point and supply voltage.
EXAMPLE 4.22
Solution
VE 110 VCC 110 (20 V) 2 V
RE RC VE
V
2V
E
1 k
IE
IC
2 mA
VRC
IC
20 V 10 V 2 V
8V
VCC VCE VE
2 mA
2 mA
IC
4 k
IB RB IC
2 mA
13.33 A
150
VRB
IB
20 V 0.7 V 2 V
VCC VBE VE
13.33 A
IB
1.3 M
Design of a Current-Gain-Stabilized
(Beta-Independent) Circuit
The circuit of Fig. 4.51 provides stabilization both for leakage and current gain (beta)
changes. The four resistor values shown must be obtained for the specified operating
point. Engineering judgment in selecting a value of emitter voltage, VE, as in the previous design consideration, leads to a direct straightforward solution for all the resistor values. The design steps are all demonstrated in the next example.
Figure 4.51 Current-gainstabilized circuit for design
considerations.
4.8 Design Operations
179
EXAMPLE 4.23
Determine the levels of RC, RE, R1, and R2 for the network of Fig. 4.51 for the operating point indicated.
Solution
VE 110 VCC 110 (20 V) 2 V
VE
V
2V
E
200 IE
IC
10 mA
VRC
V VCE VE
10 V
20 V 8 V 2 V
RC CC
IC
10 mA
IC
10 mA
RE 1 k
VB VBE VE 0.7 V 2 V 2.7 V
The equations for the calculation of the base resistors R1 and R2 will require a little thought. Using the value of base voltage calculated above and the value of the supply voltage will provide one equation—but there are two unknowns, R1 and R2. An
additional equation can be obtained from an understanding of the operation of these
two resistors in providing the necessary base voltage. For the circuit to operate efficiently, it is assumed that the current through R1 and R2 should be approximately
equal and much larger than the base current (at least 101). This fact and the voltagedivider equation for the base voltage provide the two relationships necessary to determine the base resistors. That is,
and
RE
R2 1
10
VB R2
VCC
R1 R2
Substitution yields
R2 1
10
(80)(0.2 k )
1.6 k
VB 2.7 V and
2.7R1 4.32 k
(1.6 k )(20 V)
R1 1.6 k
32 k
2.7R1 27.68 k
R1 10.25 k (use 10 k )
4.9 TRANSISTOR SWITCHING
NETWORKS
The application of transistors is not limited solely to the amplification of signals.
Through proper design it can be used as a switch for computer and control applications. The network of Fig. 4.52a can be employed as an inverter in computer logic
circuitry. Note that the output voltage VC is opposite to that applied to the base or input terminal. In addition, note the absence of a dc supply connected to the base circuit. The only dc source is connected to the collector or output side and for computer
applications is typically equal to the magnitude of the “high” side of the applied
signal—in this case 5 V.
180
Chapter 4
DC Biasing—BJTs
VCC = 5 V
RC
Vi
0.82 kΩ
5V
VC
5V
VC
RB
h FE = 125
68 kΩ
0V
0V
t
t
(a)
I C (mA)
60 µA
7
I C sat = 6.1 mA
50 µA
6
40 µA
5
4
30 µA
3
20 µA
2
10 µA
1
I B = 0 µA
0
1
~ 0V
VCE sat =
2
3
~ 0 mA
I CEO =
4
5
VCC = 5 V
VCE
(b)
Figure 4.52 Transistor inverter.
Proper design for the inversion process requires that the operating point switch
from cutoff to saturation along the load line depicted in Fig. 4.52b. For our purposes
we will assume that IC ICEO 0 mA when IB 0 A (an excellent approximation
in light of improving construction techniques), as shown in Fig. 4.52b. In addition,
we will assume that VCE VCEsat 0 V rather than the typical 0.1- to 0.3-V level.
When Vi 5 V, the transistor will be “on” and the design must ensure that the
network is heavily saturated by a level of IB greater than that associated with the IB
curve appearing near the saturation level. In Fig. 4.52b, this requires that IB 50 A.
The saturation level for the collector current for the circuit of Fig. 4.52a is defined by
ICsat VCC
RC
(4.45)
4.9 Transistor Switching Networks
181
The level of IB in the active region just before saturation results can be approximated by the following equation:
ICsat
IBmax dc
For the saturation level we must therefore ensure that the following condition is
satisfied:
IC
IB sat
(4.46)
dc
For the network of Fig. 4.52b, when Vi 5 V, the resulting level of IB is the following:
5 V 0.7 V
V 0.7 V
63 A
IB i
68 k
RB
VCC
5V
RC
0.82 k
ICsat and
6.1 mA
Testing Eq. (4.46) gives
IB 63 A ICsat
dc
6.1 mA
48.8 A
125
which is satisfied. Certainly, any level of IB greater than 60 A will pass through a
Q-point on the load line that is very close to the vertical axis.
For Vi 0 V, IB 0 A, and since we are assuming that IC ICEO 0 mA, the
voltage drop across RC as determined by VRC ICRC 0 V, resulting in VC 5 V
for the response indicated in Fig. 4.52a.
In addition to its contribution to computer logic, the transistor can also be employed as a switch using the same extremities of the load line. At saturation, the current IC is quite high and the voltage VCE very low. The result is a resistance level between the two terminals determined by
VCEsat
Rsat ICsat
and depicted in Fig. 4.53.
C
I Csat
C
+
VCE sat
–
R ≅ 0Ω
Figure 4.53 Saturation conditions and the resulting terminal
resistance.
E
E
Using a typical average value of VCEsat such as 0.15 V gives
Rsat VCEsat 0.15 V
24.6
ICsat
6.1 mA
which is a relatively low value and 0
kilohm range.
182
Chapter 4
DC Biasing—BJTs
when placed in series with resistors in the
Figure 4.54 Cutoff conditions
and the resulting terminal resistance.
For Vi 0 V, as shown in Fig. 4.54, the cutoff condition will result in a resistance
level of the following magnitude:
VCC
5V
Rcutoff ICEO
0 mA
resulting in the open-circuit equivalence. For a typical value of ICEO 10 A, the
magnitude of the cutoff resistance is
5V
VCC
500 k
10
A
ICEO
which certainly approaches an open-circuit equivalence for many situations.
Rcutoff Determine RB and RC for the transistor inverter of Fig. 4.55 if ICsat 10 mA.
EXAMPLE 4.24
VCC = 10 V
Vi
VC
RC
10 V
10 V
10 V
VC
RB
Vi
0V
h FE = 250
0V
0V
t
t
Figure 4.55 Inverter for Example 4.24.
Solution
At saturation:
ICsat 10 mA and
so that
RC VCC
RC
10 V
RC
10 V
1k
10 mA
At saturation:
IB IC sat
10 mA
40 A
dc
250
Choosing IB 60 A to ensure saturation and using
IB Vi 0.7 V
RB
4.9 Transistor Switching Networks
183
RB we obtain
10 V 0.7 V
Vi 0.7 V
155 k
60 A
IB
Choose RB 150 k , which is a standard value. Then
Vi 0.7 V
10 V 0.7 V
62 A
150 k
RB
ICsat
40 A
IB 62 A dc
IB and
Therefore, use RB 150 k and RC 1 k.
There are transistors that are referred to as switching transistors due to the speed
with which they can switch from one voltage level to the other. In Fig. 3.23c the periods of time defined as ts, td, tr, and tf are provided versus collector current. Their
impact on the speed of response of the collector output is defined by the collector
current response of Fig. 4.56. The total time required for the transistor to switch from
the “off” to the “on” state is designated as ton and defined by
ton tr td
(4.47)
with td the delay time between the changing state of the input and the beginning of
a response at the output. The time element tr is the rise time from 10% to 90% of the
final value.
Transistor "on"
Transistor "off"
IC
100%
90%
10%
0
t
ts
td
tf
tr
t on
t off
Figure 4.56 Defining the time intervals of a pulse waveform.
The total time required for a transistor to switch from the “on” to the “off” state
is referred to as toff and is defined by
toff ts tf
(4.48)
where ts is the storage time and tf the fall time from 90% to 10% of the initial value.
184
Chapter 4
DC Biasing—BJTs
For the general-purpose transistor of Fig. 3.23c at IC 10 mA, we find that
ts 120 ns
td 25 ns
tr 13 ns
and
tf 12 ns
so that
ton tr td 13 ns 25 ns 38 ns
and
toff ts tf 120 ns 12 ns 132 ns
Comparing the values above with the following parameters of a BSV52L switching
transistor reveals one of the reasons for choosing a switching transistor when the need
arises.
ton 12 ns
and
toff 18 ns
4.10 TROUBLESHOOTING TECHNIQUES
The art of troubleshooting is such a broad topic that a full range of possibilities and
techniques cannot be covered in a few sections of a book. However, the practitioner
should be aware of a few basic maneuvers and measurements that can isolate the problem area and possibly identify a solution.
Quite obviously, the first step in being able to troubleshoot a network is to fully
understand the behavior of the network and to have some idea of the expected voltage and current levels. For the transistor in the active region, the most important measurable dc level is the base-to-emitter voltage.
For an “on” transistor, the voltage VBE should be in the neighborhood of
0.7 V.
The proper connections for measuring VBE appear in Fig. 4.57. Note that the positive (red) lead is connected to the base terminal for an npn transistor and the negative (black) lead to the emitter terminal. Any reading totally different from the expected level of about 0.7 V, such as 0, 4, or 12 V, or negative in value would be suspect
and the device or network connections should be checked. For a pnp transistor, the
same connections can be used but a negative reading should be expected.
A voltage level of equal importance is the collector-to-emitter voltage. Recall from
the general characteristics of a BJT that levels of VCE in the neighborhood of 0.3 V
suggest a saturated device—a condition that should not exist unless being employed
in a switching mode. However:
For the typical transistor amplifier in the active region, VCE is usually about
25% to 75% of VCC.
For VCC 20 V, a reading of VCE of 1 to 2 V or 18 to 20 V as measured in Fig.
4.58 is certainly an uncommon result, and unless knowingly designed for this response
the design and operation should be investigated. If VCE 20 V (with VCC 20 V) at
least two possibilities exist—either the device (BJT) is damaged and has the
Figure 4.57 Checking the dc
level of VBE.
Figure 4.58 Checking the dc
level of VCE.
4.10 Troubleshooting Techniques
185
characteristics of an open circuit between collector and emitter terminals or a connection in the collector–emitter or base–emitter circuit loop is open as shown in Fig.
4.59, establishing IC at 0 mA and VRC 0 V. In Fig. 4.59, the black lead of the voltmeter is connected to the common ground of the supply and the red lead to the bottom terminal of the resistor. The absence of a collector current and a resulting drop
across RC will result in a reading of 20 V. If the meter is connected to the collector
terminal of the BJT, the reading will be 0 V since VCC is blocked from the active device by the open circuit. One of the most common errors in the laboratory experience
is the use of the wrong resistance value for a given design. Imagine the impact of using a 680- resistor for RB rather than the design value of 680 k . For VCC 20 V
and a fixed-bias configuration, the resulting base current would be
IB Figure 4.59 Effect of a poor
connection or damaged device.
20 V 0.7 V
28.4 mA
680
Figure 4.60 Checking voltage
levels with respect to ground.
rather than the desired 28.4 A—a significant difference!
A base current of 28.4 mA would certainly place the design in a saturation region
and possibly damage the device. Since actual resistor values are often different from
the nominal color-code value (recall the common tolerance levels for resistive elements), it is time well spent to measure a resistor before inserting it in the network.
The result is actual values closer to theoretical levels and some insurance that the correct resistance value is being employed.
There are times when frustration will develop. You have checked the device on a
curve tracer or other BJT testing instrumentation and it looks good. All resistor levels seem correct, the connections appear solid, and the proper supply voltage has been
applied—what next? Now the troubleshooter must strive to attain a higher level of
sophistication. Could it be that the internal connection between the wire and the end
connection of a lead is faulty? How often has simply touching a lead at the proper
point created a “make or break” situation between connections? Perhaps the supply
was turned on and set at the proper voltage but the current-limiting knob was left in
the zero position, preventing the proper level of current as demanded by the network
design. Obviously, the more sophisticated the system, the broader the range of possibilities. In any case, one of the most effective methods of checking the operation of
a network is to check various voltage levels with respect to ground by hooking up the
black (negative) lead of a voltmeter to ground and “touching” the important terminals with the red (positive) lead. In Fig. 4.60, if the red lead is connected directly to
VCC, it should read VCC volts since the network has one common ground for the supply and network parameters. At VC the reading should be less, as determined by the
drop across RC and VE should be less than VC by the collector–emitter voltage VCE.
The failure of any of these points to register what would appear to be a reasonable
level may be sufficient in itself to define the faulty connection or element. If VRC and
VRE are reasonable values but VCE is 0 V, the possibility exists that the BJT is damaged and displays a short-circuit equivalence between collector and emitter terminals.
As noted earlier, if VCE registers a level of about 0.3 V as defined by VCE VC VE
(the difference of the two levels as measured above), the network may be in saturation with a device that may or may not be defective.
It should be somewhat obvious from the discussion above that the voltmeter section of the VOM or DMM is quite important in the troubleshooting process. Current
levels are usually calculated from the voltage levels across resistors rather than “breaking” the network to insert the milliammeter section of a multimeter. On large schematics, specific voltage levels are provided with respect to ground for easy checking and
identification of possible problem areas. Of course, for the networks covered in this
chapter, one must simply be aware of typical levels within the system as defined by
the applied potential and general operation of the network.
186
Chapter 4
DC Biasing—BJTs
All in all, the troubleshooting process is a true test of your clear understanding of
the proper behavior of a network and the ability to isolate problem areas using a few
basic measurements with the appropriate instruments. Experience is the key, and that
will come only with continued exposure to practical circuits.
Based on the readings provided in Fig. 4.61, determine whether the network is operating properly and, if not, the probable cause.
EXAMPLE 4.25
Figure 4.61 Network for Example 4.25.
Solution
The 20 V at the collector immediately reveals that IC 0 mA, due to an open circuit
or a nonoperating transistor. The level of VRB 19.85 V also reveals that the transistor is “off ” since the difference of VCC VRB 0.15 V is less than that required to
turn “on” the transistor and provide some voltage for VE. In fact, if we assume a short
circuit condition from base to emitter, we obtain the following current through RB:
IRB VCC
20 V
79.4 A
RB RE
252 k
which matches that obtained from
IRB VRB
19.85 V
79.4 A
RB
250 k
If the network were operating properly, the base current should be
IB VCC VBE
20 V 0.7 V
19.3 V
42.7 A
RB ( 1)RE
250 k (101)(2 k )
452 k
The result, therefore, is that the transistor is in a damaged state, with a short-circuit
condition between base and emitter.
Based on the readings appearing in Fig. 4.62, determine whether the transistor is “on”
and the network is operating properly.
4.10 Troubleshooting Techniques
EXAMPLE 4.26
187
Solution
Based on the resistor values of R1 and R2 and the magnitude of VCC, the voltage
VB 4 V seems appropriate (and in fact it is). The 3.3 V at the emitter results in a
0.7-V drop across the base-to-emitter junction of the transistor, suggesting an “on”
transistor. However, the 20 V at the collector reveals that IC 0 mA, although the
connection to the supply must be “solid” or the 20 V would not appear at the collector of the device. Two possibilities exist—there can be a poor connection between
RC and the collector terminal of the transistor or the transistor has an open base-tocollector junction. First, check the continuity at the collector junction using an ohmmeter, and if okay, the transistor should be checked using one of the methods described
in Chapter 3.
Figure 4.62 Network for
Example 4.26.
4.11 PNP TRANSISTORS
The analysis thus far has been limited totally to npn transistors to ensure that the initial analysis of the basic configurations was as clear as possible and uncomplicated
by switching between types of transistors. Fortunately, the analysis of pnp transistors
follows the same pattern established for npn transistors. The level of IB is first determined, followed by the application of the appropriate transistor relationships to determine the list of unknown quantities. In fact, the only difference between the resulting equations for a network in which an npn transistor has been replaced by a pnp
transistor is the sign associated with particular quantities.
As noted in Fig. 4.63, the double-subscript notation continues as normally defined. The current directions, however, have been reversed to reflect the actual conduction directions. Using the defined polarities of Fig. 4.63, both VBE and VCE will
be negative quantities.
Applying Kirchhoff’s voltage law to the base–emitter loop will result in the following equation for the network of Fig. 4.63:
IERE VBE IBRB VCC 0
Substituting IE ( 1)IB and solving for IB yields
IB Figure 4.63 pnp transistor in an
emitter-stabilized configuration.
VCC VBE
RB 1)RE
(4.49)
The resulting equation is the same as Eq. (4.17) except for the sign for VBE. However, in this case VBE 0.7 V and the substitution of values will result in the same
sign for each term of Eq. (4.49) as Eq. (4.17). Keep in mind that the direction of IB
is now defined opposite of that for a pnp transistor as shown in Fig. 4.63.
For VCE Kirchhoff’s voltage law is applied to the collector–emitter loop, resulting in the following equation:
IERE VCE ICRC VCC 0
Substituting IE IC gives
VCE VCC IC(RC RE)
(4.50)
The resulting equation has the same format as Eq. (4.19), but the sign in front of
each term on the right of the equal sign has changed. Since VCC will be larger than
the magnitude of the succeeding term, the voltage VCE will have a negative sign, as
noted in an earlier paragraph.
188
Chapter 4
DC Biasing—BJTs
Determine VCE for the voltage-divider bias configuration of Fig. 4.64.
EXAMPLE 4.27
–18 V
47 kΩ
2.4 kΩ
10 µF
vo
C
10 µF
B
vi
+
VCE
β = 120
–
E
10 kΩ
1.1 kΩ
Figure 4.64 pnp transistor in a
voltage-divider bias configuration.
Solution
Testing the condition
RE
results in
(120)(1.1 k )
132 k
10R2
10(10 k )
100 k
(satisfied)
Solving for VB, we have
VB (10 k )(18 V)
R2VCC
3.16 V
47 k 10 k
R1 R2
Note the similarity in format of the equation with the resulting negative voltage for
VB.
Applying Kirchhoff’s voltage law around the base–emitter loop yields
VB VBE VE 0
VE VB VBE
and
Substituting values, we obtain
VE 3.16 V (0.7 V)
3.16 V 0.7 V
2.46 V
Note in the equation above that the standard single- and double-subscript notation is
employed. For an npn transistor the equation VE VB VBE would be exactly the
same. The only difference surfaces when the values are substituted.
The current
IE VE
2.46 V
2.24 mA
RE
1.1 k
For the collector–emitter loop:
IERE VCE ICRC VCC 0
Substituting IE IC and gathering terms, we have
VCE VCC IC(RC RE)
4.11 PNP Transistors
189
Substituting values gives
VCE 18 V (2.24 mA)(2.4 k
1.1 k )
18 V 7.84 V
10.16 V
4.12 BIAS STABILIZATION
The stability of a system is a measure of the sensitivity of a network to variations in
its parameters. In any amplifier employing a transistor the collector current IC is sensitive to each of the following parameters:
: increases with increase in temperature
VBE: decreases about 7.5 mV per degree Celsius (°C) increase in temperature
ICO (reverse saturation current): doubles in value for every 10°C increase in
temperature
Any or all of these factors can cause the bias point to drift from the designed point
of operation. Table 4.1 reveals how the level of ICO and VBE changed with increase
in temperature for a particular transistor. At room temperature (about 25°C) ICO 0.1 nA, while at 100°C (boiling point of water) ICO is about 200 times larger at 20
nA. For the same temperature variation, increased from 50 to 80 and VBE dropped
from 0.65 to 0.48 V. Recall that IB is quite sensitive to the level of VBE, especially for
levels beyond the threshold value.
TABLE 4.1 Variation of Silicon Transistor
Parameters with Temperature
T (°C)
65
25
100
175
ICO (nA)
0.2 103
0.1
20
3.3 103
20
50
80
120
VBE(V)
0.85
0.65
0.48
0.30
The effect of changes in leakage current (ICO) and current gain () on the dc bias
point is demonstrated by the common-emitter collector characteristics of Fig. 4.65a
and b. Figure 4.65 shows how the transistor collector characteristics change from a
temperature of 25°C to a temperature of 100°C. Note that the significant increase in
leakage current not only causes the curves to rise but also an increase in beta, as revealed by the larger spacing between curves.
An operating point may be specified by drawing the circuit dc load line on the
graph of the collector characteristic and noting the intersection of the load line and
the dc base current set by the input circuit. An arbitrary point is marked in Fig. 4.65a
at IB 30 A. Since the fixed-bias circuit provides a base current whose value depends approximately on the supply voltage and base resistor, neither of which is affected by temperature or the change in leakage current or beta, the same base current
magnitude will exist at high temperatures as indicated on the graph of Fig. 4.65b. As
the figure shows, this will result in the dc bias point’s shifting to a higher collector
current and a lower collector–emitter voltage operating point. In the extreme, the transistor could be driven into saturation. In any case, the new operating point may not
190
Chapter 4
DC Biasing—BJTs
Figure 4.65 Shift in dc bias point (Q-point) due to change in temperature:
(a) 25°C; (b) 100°C.
be at all satisfactory, and considerable distortion may result because of the bias-point
shift. A better bias circuit is one that will stabilize or maintain the dc bias initially
set, so that the amplifier can be used in a changing-temperature environment.
Stability Factors, S(ICO), S(VBE), and S()
A stability factor, S, is defined for each of the parameters affecting bias stability as
listed below:
S(ICO) IC
ICO
(4.51)
S(VBE) IC
VBE
(4.52)
IC
(4.53)
S() In each case, the delta symbol () signifies change in that quantity. The numerator of
each equation is the change in collector current as established by the change in the
quantity in the denominator. For a particular configuration, if a change in ICO fails to
produce a significant change in IC, the stability factor defined by S(ICO) IC/ICO
will be quite small. In other words:
Networks that are quite stable and relatively insensitive to temperature variations have low stability factors.
In some ways it would seem more appropriate to consider the quantities defined
by Eqs. (4.51–4.53) to be sensitivity factors because:
4.12 Bias Stabilization
191
The higher the stability factor, the more sensitive the network to variations in
that parameter.
The study of stability factors requires the knowledge of differential calculus. Our
purpose here, however, is to review the results of the mathematical analysis and to
form an overall assessment of the stability factors for a few of the most popular bias
configurations. A great deal of literature is available on this subject, and if time permits, you are encouraged to read more on the subject.
S(ICO): EMITTER-BIAS CONFIGURATION
For the emitter-bias configuration, an analysis of the network will result in
S(ICO) ( 1)
1 RB/RE
( 1) RB/RE
(4.54)
For RB/RE ( 1), Eq. (4.54) will reduce to the following:
S(ICO) 1
(4.55)
as shown on the graph of S(ICO) versus RB/RE in Fig. 4.66.
Figure 4.66 Variation of stability factor S(ICO) with the resistor ratio RB /RE for the emitter-bias configuration.
For RB/RE 1, Eq. (4.54) will approach the following level (as shown in Fig.
4.66):
S(ICO) ( 1)
1
→1
( 1)
(4.56)
revealing that the stability factor will approach its lowest level as RE becomes sufficiently large. Keep in mind, however, that good bias control normally requires that
RB be greater than RE. The result therefore is a situation where the best stability levels are associated with poor design criteria. Obviously, a trade-off must occur that
will satisfy both the stability and bias specifications. It is interesting to note in Fig.
4.66 that the lowest value of S(ICO) is 1, revealing that IC will always increase at a
rate equal to or greater than ICO.
For the range where RB/RE ranges between 1 and ( 1), the stability factor will
be determined by
S(ICO) 192
Chapter 4
DC Biasing—BJTs
RB
RE
(4.57)
as shown in Fig. 4.66. The results reveal that the emitter-bias configuration is quite
stable when the ratio RB/RE is as small as possible and the least stable when the same
ratio approaches ( 1).
Calculate the stability factor and the change in IC from 25°C to 100°C for the transistor defined by Table 4.1 for the following emitter-bias arrangements.
(a) RB/RE 250 (RB 250RE)
(b) RB/RE 10 (RB 10RE).
(c) RB/RE 0.01 (RE 100RB).
EXAMPLE 4.28
Solution
RB/RE
1 RB/RE
251
1 250
51
51
301
51 250
42.53
which begins to approach the level defined by 1 51.
(a) S(ICO) ( 1) 1 IC [S(ICO)](ICO) (42.53)(19.9 nA)
0.85 A
1 RB/RE
(b) S(ICO) ( 1)
1 RB/RE
11
1 10
51
51
61
51 10
9.2
IC [S(ICO)]( ICO) (9.2)(19.9 nA)
0.18 A
1 RB/RE
(c) S(ICO) ( 1)
1 RB/RE
1 0.01
1.01
51
51
51 0.01
51.01
1.01
which is certainly very close to the level of 1 forecast if RB/RE 1.
IC [S(ICO)]( ICO) 1.01(19.9 nA)
20.1 nA
Example 4.28 reveals how lower and lower levels of ICO for the modern-day BJT
transistor have improved the stability level of the basic bias configurations. Even
though the change in IC is considerably different in a circuit having ideal stability
(S 1) from one having a stability factor of 42.53, the change in IC is not that significant. For example, the amount of change in IC from a dc bias current set at, say,
2 mA, would be from 2 to 2.085 mA in the worst case, which is obviously small
enough to be ignored for most applications. Some power transistors exhibit larger
leakage currents, but for most amplifier circuits the lower levels of ICO have had a
very positive impact on the stability question.
FIXED-BIAS CONFIGURATION
For the fixed-bias configuration, if we multiply the top and bottom of Eq. (4.54)
by RE and then plug in RE 0 , the following equation will result:
S(ICO) 1
(4.58)
4.12 Bias Stabilization
193
Note that the resulting equation matches the maximum value for the emitter-bias
configuration. The result is a configuration with a poor stability factor and a high sensitivity to variations in ICO.
Voltage-Divider Bias Configuration
Recall from Section 4.5 the development of the Thévenin equivalent network appearing in Fig. 4.67, for the voltage-divider bias configuration. For the network of
Fig. 4.67, the equation for S(ICO) is the following:
S(ICO) ( 1)
Figure 4.67 Equivalent circuit
for the voltage-divider configuration.
1 RTh/RE
( 1) RTh/RE
(4.59)
Note the similarities with Eq. (4.54), where it was determined that S(ICO) had its
lowest level and the network had its greatest stability when RE RB. For Eq. (4.59),
the corresponding condition is RE RTh or RTh/RE should be as small as possible.
For the voltage-divider bias configuration, RTh can be much less than the corresponding RB of the emitter-bias configuration and still have an effective design.
Feedback-Bias Configuration (RE 5 0 )
In this case,
S(ICO) ( 1)
1 RB/RC
( 1) RB/RC
(4.60)
Since the equation is similar in format to that obtained for the emitter-bias and voltage-divider bias configurations, the same conclusions regarding the ratio RB/RC can
be applied here also.
Physical Impact
Equations of the type developed above often fail to provide a physical sense for why
the networks perform as they do. We are now aware of the relative levels of stability
and how the choice of parameters can affect the sensitivity of the network, but without the equations it may be difficult for us to explain in words why one network is
more stable than another. The next few paragraphs attempt to fill this void through
the use of some of the very basic relationships associated with each configuration.
For the fixed-bias configuration of Fig. 4.68a, the equation for the base current is
the following:
IB VCC VBE
RB
with the collector current determined by
IC IB ( 1)ICO
(4.61)
If IC as defined by Eq. (4.61) should increase due to an increase in ICO, there is
nothing in the equation for IB that would attempt to offset this undesirable increase
in current level (assuming VBE remains constant). In other words, the level of IC would
continue to rise with temperature, with IB maintaining a fairly constant value—a very
unstable situation.
For the emitter-bias configuration of Fig. 4.68b, however, an increase in IC due
to an increase in ICO will cause the voltage VE IERE ICRE to increase. The result
is a drop in the level of IB as determined by the following equation:
194
Chapter 4
DC Biasing—BJTs
Figure 4.68 Review of biasing
managements and the stability
factor S(ICO).
IB ↓ VCC VBE VE ↑
RB
(4.62)
A drop in IB will have the effect of reducing the level of IC through transistor action and thereby offset the tendency of IC to increase due to an increase in temperature. In total, therefore, the configuration is such that there is a reaction to an increase
in IC that will tend to oppose the change in bias conditions.
The feedback configuration of Fig. 4.68c operates in much the same way as the
emitter-bias configuration when it comes to levels of stability. If IC should increase
due to an increase in temperature, the level of VRC will increase in the following equation:
VCC VBE VRC ↑
(4.63)
IB ↓ RB
and the level of IB will decrease. The result is a stabilizing effect as described for the
emitter-bias configuration. One must be aware that the action described above does
not happen in a step-by-step sequence. Rather, it is a simultaneous action to maintain
the established bias conditions. In other words, the very instant IC begins to rise
the network will sense the change and the balancing effect described above will take
place.
The most stable of the configurations is the voltage-divider bias network of Fig.
4.68d. If the condition RE 10R2 is satisfied, the voltage VB will remain fairly constant for changing levels of IC. The base-to-emitter voltage of the configuration is determined by VBE VB VE. If IC should increase, VE will increase as described
above, and for a constant VB the voltage VBE will drop. A drop in VBE will establish
a lower level of IB, which will try to offset the increased level of IC.
S(VBE)
The stability factor defined by
S(VBE) IC
VBE
will result in the following equation for the emitter-bias configuration:
S(VBE) Substituting RE 0
RB ( 1)RE
(4.64)
as occurs for the fixed-bias configuration will result in
S(VBE) RB
(4.65)
4.12 Bias Stabilization
195
Equation (4.64) can be written in the following form:
S(VBE) /RE
RB/RE ( 1)
(4.66)
Substituting the condition ( 1) RB/RE will result in the following equation
for S(VBE):
S(VBE) /RE
/RE
1
1
RE
(4.67)
revealing that the larger the resistance RE, the lower the stability factor and the more
stable the system.
EXAMPLE 4.29
Determine the stability factor S(VBE) and the change in IC from 25°C to 100°C for
the transistor defined by Table 4.1 for the following bias arrangements.
(a) Fixed-bias with RB 240 k and 100.
(b) Emitter-bias with RB 240 k , RE 1 k , and 100.
(c) Emitter-bias with RB 47 k , RE 4.7 k , and 100.
Solution
RB
100
240 k
(a) Eq. (4.65): S(VBE) 0.417 103
IC [S(VBE)](VBE)
and
(0.417 103)(0.48 V 0.65 V)
(0.417 103)(0.17 V)
70.9 A
(b) In this case, ( 1) 101 and RB/RE 240. The condition ( 1) RB/RE is
not satisfied, negating the use of Eq. (4.67) and requiring the use of Eq. (4.64).
Eq. (4.64):
S(VBE) RB ( 1)RE
240 k
100
(101)1 k
100
341 k
0.293 103
which is about 30% less than the fixed-bias value due to the additional ( 1)RE
term in the denominator of the S(VBE) equation.
IC [S(VBE)](VBE)
(0.293103)(0.17 V)
50 A
(c) In this case,
( 1) 101 196
Chapter 4
DC Biasing—BJTs
RB
47 k
RE
4.7 k
10 (satisfied)
Eq. (4.67):
S(VBE) 1
RE
1
4.7 k
0.212 103
IC [S(VBE)](VBE)
and
(0.212 103)(0.17 V)
36.04 A
In Example 4.29, the increase of 70.9 A will have some impact on the level of
ICQ. For a situation where ICQ 2 mA, the resulting collector current will increase to
ICQ 2 mA 70.9 A
2.0709 mA
a 3.5% increase.
For the voltage-divider configuration, the level of RB will be changed to RTh in Eq.
(4.64) (as defined by Fig. 4.67). In Example 4.29, the use of RB 47 k is a questionable design. However, RTh for the voltage-divider configuration can be this level or
lower and still maintain good design characteristics. The resulting equation for S(VBE)
for the feedback network will be similar to that of Eq. (4.64) with RE replaced by RC.
S()
The last stability factor to be investigated is that of S(). The mathematical development is more complex than that encountered for S(ICO) and S(VBE), as suggested
by the following equation for the emitter-bias configuration:
S() IC1(1 RB/RE)
IC
1(1 2 RB/RE)
(4.68)
The notation IC1 and 1 is used to define their values under one set of network
conditions, while the notation 2 is used to define the new value of beta as established by such causes as temperature change, variation in for the same transistor,
or a change in transistors.
Determine ICQ at a temperature of 100°C if ICQ 2 mA at 25°C. Use the transistor
described by Table 4.1, where 1 50 and 2 80, and a resistance ratio RB/RE of
20.
Solution
Eq. (4.68):
S() EXAMPLE 4.30
IC1(1 RB/RE)
1(1 2 RB/RE)
(2 103)(1 20)
42 103
(50)(1 80 20)
5050
8.32 106
and
IC [S()][]
(8.32 106)(30)
0.25 mA
4.12 Bias Stabilization
197
In conclusion therefore the collector current changed from 2 mA at room temperature to 2.25 mA at 100°C, representing a change of 12.5%.
The fixed-bias configuration is defined by S() IC1/1 and RB of Eq. (4.68) can
be replaced by RTh for the voltage-divider configuration.
For the collector feedback configuration with RE 0 ,
S() IC1(RB RC)
1(RB RC(1 2))
(4.69)
Summary
Now that the three stability factors of importance have been introduced, the total effect on the collector current can be determined using the following equation:
IC S(ICO)ICO S(VBE)VBE S()
(4.70)
The equation may initially appear quite complex, but take note that each component is simply a stability factor for the configuration multiplied by the resulting change
in a parameter between the temperature limits of interest. In addition, the IC to be
determined is simply the change in IC from the level at room temperature.
For instance, if we examine the fixed-bias configuration, Eq. (4.70) becomes the
following:
IC
VBE 1 (4.71)
IC ( 1)ICO 1
RB
after substituting the stability factors as derived in this section. Let us now use Table
4.1 to find the change in collector current for a temperature change from 25°C (room
temperature) to 100°C (the boiling point of water). For this range the table reveals that
ICO 20 nA 0.1 nA 19.9 nA
VBE 0.48 V 0.65 V 0.17 V (note the sign)
80 50 30
and
Starting with a collector current of 2 mA with an RB of 240 k , the resulting
change in IC due to an increase in temperature of 75°C is the following:
IC (50 1)(19.9 nA) 2 mA
50
(0.17 V) (30)
50
240 k
1.01 A 35.42 A 1200 A
1.236 mA
which is a significant change due primarily to the change in . The collector current
has increased from 2 to 3.236 mA—but this was expected in the sense that we recognize from the content of this section that the fixed-bias configuration is the least
stable.
If the more stable voltage-divider configuration were employed with a ratio
RTh/RE 2 and RE 4.7 k , then
S(ICO) 2.89,
and
S(VBE) 0.2 103,
IC (2.89)(19.9 nA) 0.2 103(0.17 V) 1.445 106(30)
57.51 nA 34 A 43.4 A
0.077 mA
198
S() 1.445 106
Chapter 4
DC Biasing—BJTs
The resulting collector current is 2.077 mA, or essentially 2.1 mA, compared to
the 2.0 mA at 25°C. The network is obviously a great deal more stable than the fixedbias configuration, as mentioned in earlier discussions. In this case, S() did not override the other two factors and the effects of S(VBE) and S(ICO) were equally important. In fact, at higher temperatures, the effects of S(ICO) and S(VBE) will be greater
than S() for the device of Table 4.1. For temperatures below 25°C, IC will decrease
with increasingly negative temperature levels.
The effect of S(ICO) in the design process is becoming a lesser concern because
of improved manufacturing techniques that continue to lower the level of ICO ICBO.
It should also be mentioned that for a particular transistor the variation in levels of
ICBO and VBE from one transistor to another in a lot is almost negligible compared to
the variation in beta. In addition, the results of the analysis above support the fact that
for a good stabilized design:
The ratio RB/RE or RTh/RE should be as small as possible with due consideration to all aspects of the design, including the ac response.
Although the analysis above may have been clouded by some of the complex
equations for some of the sensitivities, the purpose here was to develop a higher level
of awareness of the factors that go into a good design and to be more intimate with
the transistor parameters and their impact on the network’s performance. The analysis of the earlier sections was for idealized situations with nonvarying parameter values. We are now more aware of how the dc response of the design can vary with the
parameter variations of a transistor.
4.13 PSPICE WINDOWS
Voltage-Divider Configuration
The results of Example 4.7 will now be verified using PSpice Windows. Using methods described in previous chapters, the network of Fig. 4.69 can be constructed. Recall that the transistor can be found in the EVAL.slb library, the dc source under
SOURCE.slb, and the resistor under ANALOG.slb. The capacitor will also appear
in the ANALOG.slb library. Three VIEWPOINTS appear in Fig. 4.69 as obtained
from the SPECIAL.slb library. The collector current will be sensed by the IPROBE
option, also appearing in the SPECIAL.slb library. Recall that a positive result is obtained for IPROBE if the direction of conventional current enters that side of the
symbol with the internal curve representing the scale of the meter. We will want to
set the value of beta for the transistor to match that of the example. This is accom-
Figure 4.69 Applying
PSpice Windows to the
voltage-divider configuration of Example 4.7.
4.13 PSpice Windows
199
plished by clicking on the transistor symbol (to obtain the red outline) followed by
Edit-Model-Edit Instance Model (text) to obtain the Model Editor. Then Bf is
changed to 140 to match the value of Example 4.7. Click OK, and the network is set
up for the analysis.
In this case, since we are only interested in the dc response, the Probe Setup under Analysis should enable Do not auto-run Probe. It will save us from having to
deal with the Probe response before viewing the output file or screen. The sequence
Analysis-Simulate will result in the dc levels appearing in Fig. 4.69, which closely
match those of Example 4.7. The collector-to-emitter voltage is 13.76 V 1.259 V
12.5 V, versus 12.22 V of Example 4.7, and the collector current is 0.824 mA, versus 0.85 mA. Any differences are due to the fact that we are using an actual transistor with a host of parameters not considered in our analysis. Recall the difference in
beta from the specification value and the value obtained from the plot of the previous chapter.
Since the voltage-divider network is one that is to have a low sensitivity to changes
in beta, let us return to the transistor and replace the beta of 140 with the default value
of 225.9 and examine the results. The analysis will result in the dc levels appearing
in Fig. 4.70, which are very close to those of Fig. 4.69.
Figure 4.70 Response obtained
after changing from 140 to 255.9
for the network of
Figure 4.69.
The collector-to-emitter voltage is 13.69 V 1.266 V 12.42 V, which is very
close to that obtained with a much lower beta. The collector current is actually closer
to the hand-calculated level, 0.832 mA versus 0.85 mA. There is no question, therefore, that the voltage-divider configuration demonstrates a low sensitivity to changes
in beta. Recall, however, that the fixed-bias configuration was very sensitive to changes
in beta, and let us proceed with the same type of analysis for the fixed-bias configuration and compare notes.
Fixed-Bias Configuration
The fixed-bias configuration of Fig. 4.71 is from Example 4.1 to permit a comparison of results. Beta was set to 50 using the procedure described above. In this case,
we will use a VIEWPOINT to read the collector-to-emitter voltage and enable the
display of bias currents (using the icon with the large capital I). In addition, we will
inhibit the display of some bias currents using the icon with the smaller capital I and
the diode symbol. The final touch is to move some of the currents displayed to clean
up the presentation.
200
Chapter 4
DC Biasing—BJTs
Figure 4.71 Fixed-bias configuration
with a of 50.
A PSpice analysis of the network will result in the levels appearing in Fig. 4.71.
These are a close match with the hand-written solution, with the collector voltage at
6.998 V versus 6.83 V, the collector current at 2.274 mA versus 2.35 mA, and the
base current at 47.23 A versus 47.08 A.
Let us now test the sensitivity to changes in beta by changing to the default value
of 255.9. The results appear in Fig. 4.72. Note the dramatic drop in VC to 0.113 V
compared to 6.83 V and the significant rise in ID to 5.4 mA versus the solution of
2.35 mA. The fixed-bias configuration is obviously very beta-sensitive.
Figure 4.72 Network of Figure 4.71
with a of 255.9.
4.13
PSpice Windows
201
§ 4.3 Fixed-Bias Circuit
PROBLEMS
1. For
(a)
(b)
(c)
(d)
(e)
(f)
the fixed-bias configuration of Fig. 4.73, determine:
IBQ.
ICQ.
VCEQ.
VC.
VB.
VE.
Figure 4.73 Problems 1, 4, 11,
47, 51, 52, 53
2. Given the information appearing in Fig. 4.74, determine:
(a) IC.
(b) RC.
(c) RB.
(d) VCE.
12 V
IC
RC
RB
VC = 6 V
+
VCE β = 80
I B = 40 µA
–
3. Given the information appearing in Fig. 4.75, determine:
(a) IC.
(b) VCC.
(c) .
(d) RB.
Figure 4.74 Problem 2
Figure 4.75 Problem 3
4. Find the saturation current (ICsat) for the fixed-bias configuration of Fig. 4.73.
* 5. Given the BJT transistor characteristics of Fig. 4.76:
(a) Draw a load line on the characteristics determined by E 21 V and RC 3 k for a fixedbias configuration.
(b) Choose an operating point midway between cutoff and saturation. Determine the value of
RB to establish the resulting operating point.
(c) What are the resulting values of ICQ and VCEQ?
(d) What is the value of at the operating point?
(e) What is the value of defined by the operating point?
(f) What is the saturation (ICsat) current for the design?
(g) Sketch the resulting fixed-bias configuration.
(h) What is the dc power dissipated by the device at the operating point?
(i) What is the power supplied by VCC?
(j) Determine the power dissipated by the resistive elements by taking the difference between
the results of parts (h) and (i).
202
Chapter 4
DC Biasing—BJTs
Figure 4.77 Problems 6, 9, 11,
20, 24, 48, 51, 54
Figure 4.76 Problems 5, 10, 19, 35, 36
§ 4.4 Emitter-Stabilized Bias Circuit
6. For
(a)
(b)
(c)
(d)
(e)
(f)
the emitter-stabilized bias circuit of Fig. 4.77, determine:
IBQ.
ICQ.
VCEQ.
VC.
VB.
VE.
Figure 4.78 Problem 7
7. Given the information provided in Fig. 4.78, determine:
(a) RC.
(b) RE.
(c) RB.
(d) VCE.
(e) VB.
8. Given the information provided in Fig. 4.79, determine:
(a) .
(b) VCC.
(c) RB.
9. Determine the saturation current (ICsat) for the network of Fig. 4.77.
* 10. Using the characteristics of Fig. 4.76, determine the following for an emitter-bias configuration if a Q-point is defined at ICQ 4 mA and VCEQ 10 V.
(a) RC if VCC 24 V and RE 1.2 k .
(b) at the operating point.
(c) RB.
(d) Power dissipated by the transistor.
(e) Power dissipated by the resistor RC.
Problems
Figure 4.79 Problem 8
203
* 11. (a) Determine IC and VCE for the network of Fig. 4.73.
(b) Change to 135 and determine the new value of IC and VCE for the network of Fig. 4.73.
(c) Determine the magnitude of the percent change in IC and VCE using the following equations:
%IC IC(part b) IC(part a)
IC(part a)
100%,
%VCE VCE(part b) VCE(part a)
100%
VCE(part a)
(d) Determine IC and VCE for the network of Fig. 4.77.
(e) Change to 150 and determine the new value of IC and VCE for the network of Fig. 4.77.
(f) Determine the magnitude of the percent change in IC and VCE using the following equations:
%IC IC(part e) IC(part d)
100%,
IC(part d)
%VCE VCE(part e) VCE(part d)
100%
VCE(part d)
(g) In each of the above, the magnitude of was increased 50%. Compare the percent change
in IC and VCE for each configuration, and comment on which seems to be less sensitive to
changes in .
§ 4.5 Voltage-Divider Bias
12. For
(a)
(b)
(c)
(d)
(e)
(f)
the voltage-divider bias configuration of Fig. 4.80, determine:
IBQ.
ICQ.
VCEQ.
VC.
VE.
VB.
13. Given the information provided in Fig. 4.81, determine:
(a) IC.
(b) VE.
(c) VB.
(d) R1.
14. Given the information appearing in Fig. 4.82, determine:
(a) IC.
(b) VE.
(c) VCC.
(d) VCE.
(e) VB.
(f) R1.
Figure 4.80 Problems 12, 15,
18, 20, 24, 49, 51, 52, 55
204
Figure 4.81 Problem 13
Chapter 4
DC Biasing—BJTs
Figure 4.82 Problem 14
15. Determine the saturation current (ICsat) for the network of Fig. 4.80.
* 16. Determine the following for the voltage-divider configuration of Fig. 4.83 using the approximate approach if the condition established by Eq. (4.33) is satisfied.
(a) IC.
(b) VCE.
(c) IB.
(d) VE.
(e) VB.
* 17. Repeat Problem 16 using the exact (Thévenin) approach and compare solutions. Based on the
results, is the approximate approach a valid analysis technique if Eq. (4.33) is satisfied?
18. (a) Determine ICQ, VCEQ, and IBQ for the network of Problem 12 (Fig. 4.80) using the approximate approach even though the condition established by Eq. (4.33) is not satisfied.
(b) Determine ICQ, VCEQ, and IBQ using the exact approach.
(c) Compare solutions and comment on whether the difference is sufficiently large to require
standing by Eq. (4.33) when determining which approach to employ.
* 19. (a) Using the characteristics of Fig. 4.76, determine RC and RE for a voltage-divider network
having a Q-point of ICQ 5 mA and VCEQ 8 V. Use VCC 24 V and RC 3RE.
(b) Find VE.
(c) Determine VB.
(d) Find R2 if R1 24 k assuming that RE 10R2.
(e) Calculate at the Q-point.
(f) Test Eq. (4.33), and note whether the assumption of part (d) is correct.
Figure 4.83 Problems 16, 17,
21
* 20. (a) Determine IC and VCE for the network of Fig. 4.80.
(b) Change to 120 (50% increase), and determine the new values of IC and VCE for the network of Fig. 4.80.
(c) Determine the magnitude of the percent change in IC and VCE using the following equations:
%IC IC(part b) IC(part a)
100%,
IC(part a)
%VCE VCE(part b) VCE(part a)
100%
VCE(part a)
(d) Compare the solution to part (c) with the solutions obtained for parts (c) and (f ) of Problem 11. If not performed, note the solutions provided in Appendix E.
(e) Based on the results of part (d), which configuration is least sensitive to variations in ?
* 21. (a) Repeat parts (a) through (e) of Problem 20 for the network of Fig. 4.83. Change to 180
in part (b).
(b) What general conclusions can be made about networks in which the condition RE 10R2
is satisfied and the quantities IC and VCE are to be determined in response to a change in ?
§ 4.6 DC Bias with Voltage Feedback
22. For
(a)
(b)
(c)
the collector feedback configuration of Fig. 4.84, determine:
IB.
IC.
VC.
23. For
(a)
(b)
(c)
(d)
the voltage feedback network of Fig. 4.85, determine:
IC.
VC.
VE.
VCE.
Figure 4.84 Problems 22, 50,
56
Figure 4.85 Problem 23
Problems
205
* 24. (a) Determine the level of IC and VCE for the network of Fig. 4.86.
(b) Change to 135 (50% increase), and calculate the new levels of IC and VCE.
(c) Determine the magnitude of the percent change in IC and VCE using the following equations:
%IC IC(part b) IC(part a)
100%,
IC(part a)
%VCE VCE(part b) VCE(part a)
VCE(part a)
100%
(d) Compare the results of part (c) with those of Problems 11(c), 11(f ), and 20(c). How does
the collector-feedback network stack up against the other configurations in sensitivity to
changes in ?
25. Determine the range of possible values for VC for the network of Fig. 4.87 using the 1-M
tentiometer.
* 26. Given VB 4 V for the network of Fig. 4.88, determine:
(a) VE.
(b) IC.
(c) VC.
(d) VCE.
(e) IB.
(f) .
Figure 4.87 Problem 25
Figure 4.86 Problem 24
Figure 4.88 Problem 26
§ 4.7 Miscellaneous Bias Configurations
27. Given VC 8 V for the network of Fig. 4.89, determine:
(a) IB.
(b) IC.
(c) .
(d) VCE.
* 28. For
(a)
(b)
(c)
(d)
the network of Fig. 4.90, determine:
IB.
IC.
VCE.
VC.
Figure 4.90 Problem 28
Figure 4.89 Problem 27
206
Chapter 4
DC Biasing—BJTs
po-
* 29. For
(a)
(b)
(c)
(d)
the network of Fig. 4.91, determine:
IB.
IC.
VE.
VCE.
* 30. Determine the level of VE and IE for the network of Fig. 4.92.
* 31. For the network of Fig. 4.93, determine:
(a) IE.
(b) VC.
(c) VCE.
–
2.2 kΩ
8V
–
VCE
+
VC
IE
1.8 kΩ
10 V
Figure 4.92 Problem 30
Figure 4.91 Problem 29
Figure 4.93 Problem 31
§ 4.8 Design Operations
32. Determine RC and RB for a fixed-bias configuration if VCC 12 V, 80, and ICQ 2.5 mA
with VCEQ 6 V. Use standard values.
33. Design an emitter-stabilized network at ICQ 21 ICsat and VCEQ 12 VCC. Use VCC 20 V, ICsat 10 mA, 120, and RC 4RE. Use standard values.
34. Design a voltage-divider bias network using a supply of 24 V, a transistor with a beta of 110,
and an operating point of ICQ 4 mA and VCEQ 8 V. Choose VE 18 VCC. Use standard values.
* 35. Using the characteristics of Fig. 4.76, design a voltage-divider configuration to have a saturation level of 10 mA and a Q-point one-half the distance between cutoff and saturation. The
available supply is 28 V, and VE is to be one-fifth of VCC. The condition established by Eq.
(4.33) should also be met to provide a high stability factor. Use standard values.
§ 4.9 Transistor Switching Networks
* 36. Using the characteristics of Fig. 4.76, determine the appearance of the output waveform for the
network of Fig. 4.94. Include the effects of VCEsat, and determine IB, IBmax, and ICsat when Vi 10 V. Determine the collector-to-emitter resistance at saturation and cutoff.
* 37. Design the transistor inverter of Fig. 4.95 to operate with a saturation current of 8 mA using a
transistor with a beta of 100. Use a level of IB equal to 120% of IBmax and standard resistor values.
10 V
2.4 kΩ
Vi
5V
RC
Vi
Vo
10 V
180 kΩ
Vo
5V
Vi
RB
Vi
0V
t
Figure 4.94 Problem 36
β = 100
0V
t
Figure 4.95 Problem 37
Problems
207
38. (a) Using the characteristics of Fig. 3.23c, determine ton and toff at a current of 2 mA. Note
the use of log scales and the possible need to refer to Section 11.2.
(b) Repeat part (a) at a current of 10 mA. How have ton and toff changed with increase in collector current?
(c) For parts (a) and (b), sketch the pulse waveform of Fig. 4.56 and compare results.
§ 4.10 Troubleshooting Techniques
* 39. The measurements of Fig. 4.96 all reveal that the network is not functioning correctly. List as
many reasons as you can for the measurements obtained.
Figure 4.96 Problem 39
* 40. The measurements appearing in Fig. 4.97 reveal that the networks are not operating properly.
Be specific in describing why the levels obtained reflect a problem with the expected network
behavior. In other words, the levels obtained reflect a very specific problem in each case.
16 V
16 V
3.6 kΩ
91 kΩ
VB = 9.4 V
β = 100
3.6 kΩ
91 kΩ
β = 100
2.64 V
4V
18 kΩ
1.2 kΩ
(a)
18 kΩ
1.2 kΩ
(b)
Figure 4.97 Problem 40
Figure 4.98 Problem 41
41. For
(a)
(b)
(c)
(d)
(e)
the circuit of Fig. 4.98:
Does VC increase or decrease if RB is increased?
Does IC increase or decrease if is reduced?
What happens to the saturation current if is increased?
Does the collector current increase or decrease if VCC is reduced?
What happens to VCE if the transistor is replaced by one with smaller ?
208
Chapter 4
DC Biasing—BJTs
42. Answer the following questions about the circuit of Fig. 4.99.
(a) What happens to the voltage VC if the transistor is replaced by one having a larger value
of ?
(b) What happens to the voltage VCE if the ground leg of resistor RB2 opens (does not connect
to ground)?
(c) What happens to IC if the supply voltage is low?
(d) What voltage VCE would occur if the transistor base–emitter junction fails by becoming
open?
(e) What voltage VCE would result if the transistor base–emitter junction fails by becoming a
short?
Figure 4.99 Problem 42
* 43. Answer the following questions about the circuit of Fig. 4.100.
(a) What happens to the voltage VC if the resistor RB is open?
(b) What should happen to VCE if increases due to temperature?
(c) How will VE be affected when replacing the collector resistor with one whose resistance
is at the lower end of the tolerance range?
(d) If the transistor collector connection becomes open, what will happen to VE?
(e) What might cause VCE to become nearly 18 V?
Figure 4.100 Problem 43
§ 4.11 PNP Transistors
44. Determine VC, VCE, and IC for the network of Fig. 4.101.
45. Determine VC and IB for the network of Fig. 4.102.
46. Determine IE and VC for the network of Fig. 4.103.
Figure 4.101 Problem 44
Figure 4.102 Problem 45
Figure 4.103 Problem 46
Problems
209
§ 4.12 Bias Stabilization
47. Determine the following for the network of Fig. 4.73.
(a) S(ICO).
(b) S(VBE).
(c) S() using T1 as the temperature at which the parameter values are specified and (T2) as
25% more than (T1).
(d) Determine the net change in IC if a change in operating conditions results in ICO increasing from 0.2 to 10 A, VBE drops from 0.7 to 0.5 V, and increases 25%.
* 48. For
(a)
(b)
(c)
the network of Fig. 4.77, determine:
S(ICO).
S(VBE).
S() using T1 as the temperature at which the parameter values are specified and (T2) as
25% more than (T1).
(d) Determine the net change in IC if a change in operating conditions results in ICO increasing from 0.2 to 10 A, VBE drops from 0.7 to 0.5 V, and increases 25%.
* 49. For
(a)
(b)
(c)
the network of Fig. 4.80, determine:
S(ICO).
S(VBE).
S() using T1 as the temperature at which the parameter values are specified and (T2) as
25% more than (T1).
(d) Determine the net change in IC if a change in operating conditions results in ICO increasing from 0.2 to 10 A, VBE drops from 0.7 to 0.5 V, and increases 25%.
* 50. For
(a)
(b)
(c)
the network of Fig. 4.89, determine:
S(ICO).
S(VBE).
S() using T1 as the temperature at which the parameter values are specified and (T2) as
25% more than (T1).
(d) Determine the net change in IC if a change in operating conditions results in ICO increasing from 0.2 to 10 A, VBE drops from 0.7 to 0.5 V, and increases 25%.
* 51. Compare the relative values of stability for Problems 47 through 50. The results for Exercises
47 and 49 can be found in Appendix E. Can any general conclusions be derived from the results?
* 52. (a) Compare the levels of stability for the fixed-bias configuration of Problem 47.
(b) Compare the levels of stability for the voltage-divider configuration of Problem 49.
(c) Which factors of parts (a) and (b) seem to have the most influence on the stability of the
system, or is there no general pattern to the results?
§ 4.13 PSpice Windows
53. Perform a PSpice analysis of the network of Fig. 4.73. That is, determine IC, VCE, band IB.
54. Repeat Problem 53 for the network of Fig. 4.77.
55. Repeat Problem 53 for the network of Fig. 4.80.
56. Repeat Problem 53 for the network of Fig. 4.84.
*Please Note: Asterisks indicate more difficult problems.
210
Chapter 4
DC Biasing—BJTs
CHAPTER
Field-Effect
Transistors
5
5.1 INTRODUCTION
The field-effect transistor (FET) is a three-terminal device used for a variety of applications that match, to a large extent, those of the BJT transistor described in Chapters 3 and 4. Although there are important differences between the two types of devices, there are also many similarities that will be pointed out in the sections to follow.
The primary difference between the two types of transistors is the fact that the
BJT transistor is a current-controlled device as depicted in Fig. 5.1a, while the JFET
transistor is a voltage-controlled device as shown in Fig. 5.1b. In other words, the current IC in Fig. 5.1a is a direct function of the level of IB. For the FET the current I
will be a function of the voltage VGS applied to the input circuit as shown in Fig. 5.1b.
In each case the current of the output circuit is being controlled by a parameter of the
input circuit—in one case a current level and in the other an applied voltage.
Figure 5.1 (a) Current-controlled and (b) voltage-controlled
amplifiers.
Just as there are npn and pnp bipolar transistors, there are n-channel and p-channel field-effect transistors. However, it is important to keep in mind that the BJT transistor is a bipolar device—the prefix bi- revealing that the conduction level is a function of two charge carriers, electrons and holes. The FET is a unipolar device
depending solely on either electron (n-channel) or hole (p-channel) conduction.
The term field-effect in the chosen name deserves some explanation. We are all
familiar with the ability of a permanent magnet to draw metal filings to the magnet
without the need for actual contact. The magnetic field of the permanent magnet has
enveloped the filings and attracted them to the magnet through an effort on the part
of the magnetic flux lines to be as short as possible. For the FET an electric field is
established by the charges present that will control the conduction path of the output
211
Drs. Ian Munro Ross (front) and
G. C. Dacey jointly developed an
experimental procedure for measuring
the characteristics of a field-effect
transistor in 1955. (Courtesy of AT&T
Archives.)
Dr. Ross Born: Southport, England
PhD Gonville and Caius
College, Cambridge
University President emeritus of AT&T Bell Labs Fellow—IEEE, Member of the
National Science Board
Chairman—National
Advisory Committee on
Semiconductors
Dr. Dacey Born: Chicago, Illinois
PhD California Institute
of Technology
Director of Solid-State
Electronics Research at
Bell Labs
Vice President, Research
at Sandia Corporation
Member IRE, Tau Beta Pi,
Eta Kappa Nu
212
circuit without the need for direct contact between the controlling and controlled
quantities.
There is a natural tendency when introducing a second device with a range of applications similar to one already introduced to compare some of the general characteristics of one versus the other. One of the most important characteristics of the FET
is its high input impedance. At a level of 1 to several hundred megohms it far exceeds
the typical input resistance levels of the BJT transistor configurations—a very important characteristic in the design of linear ac amplifier systems. On the other hand,
the BJT transistor has a much higher sensitivity to changes in the applied signal. In
other words, the variation in output current is typically a great deal more for BJTs
than FETs for the same change in applied voltage. For this reason, typical ac voltage
gains for BJT amplifiers are a great deal more than for FETs. In general, FETs are
more temperature stable than BJTs, and FETs are usually smaller in construction than
BJTs, making them particularly useful in integrated-circuit (IC) chips. The construction characteristics of some FETs, however, can make them more sensitive to handling than BJTs.
Two types of FETs will be introduced in this chapter: the junction field-effect
transistor (JFET) and the metal-oxide-semiconductor field-effect transistor (MOSFET). The MOSFET category is further broken down into depletion and enhancement
types, which are both described. The MOSFET transistor has become one of the most
important devices used in the design and construction of integrated circuits for digital computers. Its thermal stability and other general characteristics make it extremely popular in computer circuit design. However, as a discrete element in a typical top-hat container, it must be handled with care (to be discussed in a later
section).
Once the FET construction and characteristics have been introduced, the biasing
arrangements will be covered in Chapter 6. The analysis performed in Chapter 4 using BJT transistors will prove helpful in the derivation of the important equations and
understanding the results obtained for FET circuits.
5.2 CONSTRUCTION AND
CHARACTERISTICS OF JFETs
As indicated earlier, the JFET is a three-terminal device with one terminal capable of
controlling the current between the other two. In our discussion of the BJT transistor
the npn transistor was employed through the major part of the analysis and design
sections, with a section devoted to the impact of using a pnp transistor. For the JFET
transistor the n-channel device will appear as the prominent device, with paragraphs
and sections devoted to the impact of using a p-channel JFET.
The basic construction of the n-channel JFET is shown in Fig. 5.2. Note that the
major part of the structure is the n-type material that forms the channel between the
embedded layers of p-type material. The top of the n-type channel is connected through
an ohmic contact to a terminal referred to as the drain (D), while the lower end of
the same material is connected through an ohmic contact to a terminal referred to as
the source (S). The two p-type materials are connected together and to the gate (G)
terminal. In essence, therefore, the drain and source are connected to the ends of the
n-type channel and the gate to the two layers of p-type material. In the absence of
any applied potentials the JFET has two p-n junctions under no-bias conditions. The
result is a depletion region at each junction as shown in Fig. 5.2 that resembles the
same region of a diode under no-bias conditions. Recall also that a depletion region
is that region void of free carriers and therefore unable to support conduction through
the region.
Chapter 5
Field-Effect Transistors
Figure 5.2 Junction field-effect
transistor (JFET).
Analogies are seldom perfect and at times can be misleading, but the water analogy of Fig. 5.3 does provide a sense for the JFET control at the gate terminal and the
appropriateness of the terminology applied to the terminals of the device. The source
of water pressure can be likened to the applied voltage from drain to source that will
establish a flow of water (electrons) from the spigot (source). The “gate,” through an
applied signal (potential), controls the flow of water (charge) to the “drain.” The drain
and source terminals are at opposite ends of the n-channel as introduced in Fig. 5.2
because the terminology is defined for electron flow.
Figure 5.3 Water analogy for
the JFET control mechanism.
VGS 0 V, VDS Some Positive Value
In Fig. 5.4, a positive voltage VDS has been applied across the channel and the gate
has been connected directly to the source to establish the condition VGS 0 V. The
result is a gate and source terminal at the same potential and a depletion region in the
low end of each p-material similar to the distribution of the no-bias conditions of Fig.
5.2. The instant the voltage VDD ( VDS) is applied, the electrons will be drawn to
the drain terminal, establishing the conventional current ID with the defined direction
of Fig. 5.4. The path of charge flow clearly reveals that the drain and source currents
are equivalent (ID IS). Under the conditions appearing in Fig. 5.4, the flow of charge
is relatively uninhibited and limited solely by the resistance of the n-channel between
drain and source.
ID
D
+
n-channel
Depletion
region
e
e
G
p
+
p
n
VDS
VDD
e
e
VG S = 0 V
–
S
IS
–
Figure 5.4 JFET in the VGS 0 V and VDS 0 V.
5.2 Construction and Characteristics of JFETs
213
Figure 5.5 Varying reverse-bias
potentials across the p-n junction
of an n-channel JFET.
Figure 5.6
214
It is important to note that the depletion region is wider near the top of both ptype materials. The reason for the change in width of the region is best described
through the help of Fig. 5.5. Assuming a uniform resistance in the n-channel, the resistance of the channel can be broken down to the divisions appearing in Fig. 5.5. The
current ID will establish the voltage levels through the channel as indicated on the
same figure. The result is that the upper region of the p-type material will be reversebiased by about 1.5 V, with the lower region only reverse-biased by 0.5 V. Recall from
the discussion of the diode operation that the greater the applied reverse bias, the
wider the depletion region—hence the distribution of the depletion region as shown
in Fig. 5.5. The fact that the p-n junction is reverse-biased for the length of the channel results in a gate current of zero amperes as shown in the same figure. The fact
that IG 0 A is an important characteristic of the JFET.
As the voltage VDS is increased from 0 to a few volts, the current will increase as
determined by Ohm’s law and the plot of ID versus VDS will appear as shown in Fig.
5.6. The relative straightness of the plot reveals that for the region of low values of
VDS, the resistance is essentially constant. As VDS increases and approaches a level
referred to as VP in Fig. 5.6, the depletion regions of Fig. 5.4 will widen, causing a
noticeable reduction in the channel width. The reduced path of conduction causes the
resistance to increase and the curve in the graph of Fig. 5.6 to occur. The more horizontal the curve, the higher the resistance, suggesting that the resistance is approaching “infinite” ohms in the horizontal region. If VDS is increased to a level where
it appears that the two depletion regions would “touch” as shown in Fig. 5.7, a condition referred to as pinch-off will result. The level of VDS that establishes this condition is referred to as the pinch-off voltage and is denoted by VP as shown in Fig.
5.6. In actuality, the term pinch-off is a misnomer in that it suggests the current ID is
pinched off and drops to 0 A. As shown in Fig. 5.6, however, this is hardly the
case — ID maintains a saturation level defined as IDSS in Fig. 5.6. In reality a very
small channel still exists, with a current of very high density. The fact that ID does
not drop off at pinch-off and maintains the saturation level indicated in Fig. 5.6 is
verified by the following fact: The absence of a drain current would remove the possibility of different potential levels through the n-channel material to establish the
varying levels of reverse bias along the p-n junction. The result would be a loss of
the depletion region distribution that caused pinch-off in the first place.
ID versus VDS for VGS 0 V.
Chapter 5
Figure 5.7 Pinch-off (VGS 0 V, VDS VP).
Field-Effect Transistors
As VDS is increased beyond VP, the region of close encounter between the two
depletion regions will increase in length along the channel, but the level of ID remains
essentially the same. In essence, therefore, once VDS VP the JFET has the characteristics of a current source. As shown in Fig. 5.8, the current is fixed at ID IDSS,
but the voltage VDS (for levels VP) is determined by the applied load.
The choice of notation IDSS is derived from the fact that it is the Drain-to-Source
current with a Short-circuit connection from gate to source. As we continue to investigate the characteristics of the device we will find that:
IDSS is the maximum drain current for a JFET and is defined by the conditions
VGS 0 V and VDS |VP|.
Note in Fig. 5.6 that VGS 0 V for the entire length of the curve. The next few
paragraphs will describe how the characteristics of Fig. 5.6 are affected by changes
in the level of VGS.
Figure 5.8 Current source
equivalent for VGS 0 V,
VDS VP.
VGS 0 V
The voltage from gate to source, denoted VGS, is the controlling voltage of the JFET.
Just as various curves for IC versus VCE were established for different levels of IB for
the BJT transistor, curves of ID versus VDS for various levels of VGS can be developed
for the JFET. For the n-channel device the controlling voltage VGS is made more and
more negative from its VGS 0 V level. In other words, the gate terminal will be set
at lower and lower potential levels as compared to the source.
In Fig. 5.9 a negative voltage of 1 V has been applied between the gate and
source terminals for a low level of VDS. The effect of the applied negative-bias VGS
is to establish depletion regions similar to those obtained with VGS 0 V but at lower
levels of VDS. Therefore, the result of applying a negative bias to the gate is to reach
the saturation level at a lower level of VDS as shown in Fig. 5.10 for VGS 1 V.
The resulting saturation level for ID has been reduced and in fact will continue to decrease as VGS is made more and more negative. Note also on Fig. 5.10 how the pinchoff voltage continues to drop in a parabolic manner as VGS becomes more and more
negative. Eventually, VGS when VGS VP will be sufficiently negative to establish
a saturation level that is essentially 0 mA, and for all practical purposes the device
has been “turned off.” In summary:
ID
D
G
+
IG = 0 A
p
p
n
>
0V
–
+
VDS
1V
+
VG S = –1 V
–
S
IS
–
Figure 5.9 Application of a
negative voltage to the gate of a
JFET.
5.2 Construction and Characteristics of JFETs
215
ID (mA)
Locus of pinch-off values
Ohmic
Region
Saturation Region
VGS = 0 V
IDSS 8
7
6
5
VGS = –1 V
4
3
VGS = –2 V
2
VGS = –3 V
1
VGS = – 4 V = VP
0
10
5
VP (for VGS = 0 V)
15
20
25
VDS (V)
Figure 5.10 n-Channel JFET characteristics with IDSS 8 mA and VP 4 V.
The level of VGS that results in ID 0 mA is defined by VGS VP, with VP
being a negative voltage for n-channel devices and a positive voltage for
p-channel JFETs.
On most specification sheets the pinch-off voltage is specified as VGS(off) rather
than VP. A specification sheet will be reviewed later in the chapter when the primary
elements of concern have been introduced. The region to the right of the pinch-off
locus of Fig. 5.10 is the region typically employed in linear amplifiers (amplifiers
with minimum distortion of the applied signal) and is commonly referred to as the
constant-current, saturation, or linear amplification region.
Voltage-Controlled Resistor
The region to the left of the pinch-off locus of Fig. 5.10 is referred to as the ohmic
or voltage-controlled resistance region. In this region the JFET can actually be employed as a variable resistor (possibly for an automatic gain control system) whose
resistance is controlled by the applied gate-to-source voltage. Note in Fig. 5.10 that
the slope of each curve and therefore the resistance of the device between drain and
source for VDS VP is a function of the applied voltage VGS. As VGS becomes more
and more negative, the slope of each curve becomes more and more horizontal,
corresponding with an increasing resistance level. The following equation will provide a good first approximation to the resistance level in terms of the applied voltage
VGS.
ro
rd (1 VGS/VP)2
(5.1)
where ro is the resistance with VGS 0 V and rd the resistance at a particular level
of VGS.
For an n-channel JFET with ro equal to 10 k (VGS 0 V, VP 6 V), Eq. (5.1)
will result in 40 k at VGS 3 V.
p-Channel Devices
The p-channel JFET is constructed in exactly the same manner as the n-channel device of Fig. 5.2, but with a reversal of the p- and n-type materials as shown in Fig. 5.11.
216
Chapter 5
Field-Effect Transistors
ID
D
+
+
+
IG = 0 A
n
+
n
p
VDS
VDD
–
+
+
+
+
G
+
VGG
–
VG S = + VGG
–
S
IS
–
Figure 5.11 p-Channel JFET.
The defined current directions are reversed, as are the actual polarities for the voltages VGS and VDS. For the p-channel device, the channel will be constricted by increasing positive voltages from gate to source and the double-subscript notation for
VDS will result in negative voltages for VDS on the characteristics of Fig. 5.12, which
has an IDSS of 6 mA and a pinch-off voltage of VGS 6 V. Do not let the minus
signs for VDS confuse you. They simply indicate that the source is at a higher potential than the drain.
Figure 5.12 p-Channel JFET characteristics with IDSS 6 mA and VP 6 V.
Note at high levels of VDS that the curves suddenly rise to levels that seem unbounded. The vertical rise is an indication that breakdown has occurred and the
current through the channel (in the same direction as normally encountered) is now
limited solely by the external circuit. Although not appearing in Fig. 5.10 for the
n-channel device, they do occur for the n-channel device if sufficient voltage is applied. This region can be avoided if the level of VDSmax is noted on the specification
sheet and the design is such that the actual level of VDS is less than this value for all
values of VGS.
5.2 Construction and Characteristics of JFETs
217
Symbols
The graphic symbols for the n-channel and p-channel JFETs are provided in Fig. 5.13.
Note that the arrow is pointing in for the n-channel device of Fig. 5.13a to represent
the direction in which IG would flow if the p-n junction were forward-biased. For the
p-channel device (Fig. 5.13b) the only difference in the symbol is the direction of the
arrow.
Figure 5.13 JFET symbols: (a)
n-channel; (b) p-channel.
Summary
A number of important parameters and relationships were introduced in this section.
A few that will surface frequently in the analysis to follow in this chapter and the
next for n-channel JFETs include the following:
The maximum current is defined as IDSS and occurs when VGS 0 V and
VDS |VP| as shown in Fig. 5.14a.
For gate-to-source voltages VGS less than (more negative than) the pinch-off
level, the drain current is 0 A (ID 0 A) as appearing in Fig. 5.14b.
For all levels of VGS between 0 V and the pinch-off level, the current ID will
range between IDSS and 0 A, respectively, as reviewed by Fig. 5.14c.
For p-channel JFETs a similar list can be developed.
D
+
G
VDD ≥ VP
ID = IDSS
+
VG S
–
VG S = 0 V
D
–
S
VG S = – VGG
G
– +
VG S
VG G
–
+
VG G
≥
(a)
≥
VG G
≥ 0V
D
G
–
VG G
+
S
VP
0 mA ≥ ID > IDSS
ID
+
VG S
–
S
(c)
Chapter 5
VDD
(b)
VP
218
ID = 0 A
Field-Effect Transistors
VDD
Figure 5.14 (a) VGS 0 V,
ID IDSS; (b) cutoff (ID 0 A)
VGS less than the pinch-off level;
(c) ID exists between 0 A and IDSS
for VGS less than or equal to 0 V
and greater than the pinch-off
level.
5.3 TRANSFER CHARACTERISTICS
Derivation
For the BJT transistor the output current IC and input controlling current IB were related by beta, which was considered constant for the analysis to be performed. In
equation form,

 control variable
↓
(5.2)
IC f(IB) IB
↑

constant
In Eq. (5.2) a linear relationship exists between IC and IB. Double the level of IB and
IC will increase by a factor of two also.
Unfortunately, this linear relationship does not exist between the output and input
quantities of a JFET. The relationship between ID and VGS is defined by Shockley’s
equation:
control variable
↓ 2
V
(5.3)
ID IDSS 1 GS
VP
↑
↑
constants
The squared term of the equation will result in a nonlinear relationship between ID and
VGS, producing a curve that grows exponentially with decreasing magnitudes of VGS.
For the dc analysis to be performed in Chapter 6, a graphical rather than mathematical approach will in general be more direct and easier to apply. The graphical approach, however, will require a plot of Eq. (5.3) to represent the device and a plot of
the network equation relating the same variables. The solution is defined by the point
of intersection of the two curves. It is important to keep in mind when applying the
graphical approach that the device characteristics will be unaffected by the network
in which the device is employed. The network equation may change along with the
intersection between the two curves, but the transfer curve defined by Eq. (5.3) is unaffected. In general, therefore:
The transfer characteristics defined by Shockley’s equation are unaffected by
the network in which the device is employed.
William Bradford Shockley
(1910–1989), co-inventor of the
first transistor and formulator of
the “field-effect” theory employed
in the development of the
transistor and FET. (Courtesy of
AT&T Archives.)
Born: London, England
PhD Harvard, 1936
Head, Transistor Physics
Department–Bell Laboratories
President, Shockley Transistor
Corp.
Poniatoff Professor of
Engineering Science at
Stanford University
Nobel Prize in physics in 1956
with Drs. Brattain and Bardeen
The transfer curve can be obtained using Shockley’s equation or from the output
characteristics of Fig. 5.10. In Fig. 5.15 two graphs are provided, with the vertical
ID (mA)
ID (mA)
10 10
9 9
IDSS
8 8
VGS = 0 V
IDSS
7 7
6 6
5 5
VGS = –1 V
4 4
3 3
VGS = –2 V
VGS = –3 V
VGS = – 4 V
2 2
1 1
VGS (V) – 4
–3
–2
–1
0 0
5
10
15
20
25
ID = 0 mA, VGS = VP
5.3
Transfer Characteristics
VDS
Figure 5.15 Obtaining the
transfer curve from the drain
characteristics.
219
scaling in milliamperes for each graph. One is a plot of ID versus VDS, while the other
is ID versus VGS. Using the drain characteristics on the right of the “y” axis, a horizontal line can be drawn from the saturation region of the curve denoted VGS 0 V
to the ID axis. The resulting current level for both graphs is IDSS. The point of intersection on the ID versus VGS curve will be as shown since the vertical axis is defined
as VGS 0 V.
In review:
When VGS 0 V, ID IDSS.
When VGS VP 4 V, the drain current is zero milliamperes, defining another
point on the transfer curve. That is:
When VGS VP, ID 0 mA.
Before continuing, it is important to realize that the drain characteristics relate
one output (or drain) quantity to another output (or drain) quantity—both axes are
defined by variables in the same region of the device characteristics. The transfer characteristics are a plot of an output (or drain) current versus an input-controlling quantity. There is therefore a direct “transfer” from input to output variables when employing the curve to the left of Fig. 5.15. If the relationship were linear, the plot of
ID versus VGS would result in a straight line between IDSS and VP. However, a parabolic curve will result because the vertical spacing between steps of VGS on the drain
characteristics of Fig. 5.15 decreases noticeably as VGS becomes more and more negative. Compare the spacing between VGS 0 V and VGS 1 V to that between
VGS 3 V and pinch-off. The change in VGS is the same, but the resulting change
in ID is quite different.
If a horizontal line is drawn from the VGS 1 V curve to the ID axis and then
extended to the other axis, another point on the transfer curve can be located. Note
that VGS 1 V on the bottom axis of the transfer curve with ID 4.5 mA. Note in
the definition of ID at VGS 0 V and 1 V that the saturation levels of ID are employed and the ohmic region ignored. Continuing with VGS 2 V and 3 V, the
transfer curve can be completed. It is the transfer curve of ID versus VGS that will receive extended use in the analysis of Chapter 6 and not the drain characteristics of
Fig. 5.15. The next few paragraphs will introduce a quick, efficient method of plotting ID versus VGS given only the levels of IDSS and VP and Shockley’s equation.
Applying Shockley’s Equation
The transfer curve of Fig. 5.15 can also be obtained directly from Shockley’s equation (5.3) given simply the values of IDSS and VP. The levels of IDSS and VP define
the limits of the curve on both axes and leave only the necessity of finding a few intermediate plot points. The validity of Eq. (5.3) as a source of the transfer curve of
Fig. 5.15 is best demonstrated by examining a few specific levels of one variable and
finding the resulting level of the other as follows:
Substituting VGS 0 V gives
Eq. (5.3):
IDSS
Chapter 5
2
P
ID IDSS | VGS 0 V
and
220
0
1 V I
Field-Effect Transistors
2
V
ID IDSS 1 GS
VP
DSS(1
0)2
(5.4)
Substituting VGS VP yields
VP
ID IDSS 1 VP
2
IDSS(1 1)2 IDSS (0)
ID 0 AVGS VP
(5.5)
For the drain characteristics of Fig. 5.15, if we substitute VGS 1 V,
V
ID IDSS 1 GS
VP
2
1 V
8 mA 1 4 V
8 mA(0.5625)
4.5 mA
2
2
1
8 mA 1 4
8 mA(0.75)2
as shown in Fig. 5.15. Note the care taken with the negative signs for VGS and VP in
the calculations above. The loss of one sign would result in a totally erroneous result.
It should be obvious from the above that given IDSS and VP (as is normally provided on specification sheets) the level of ID can be found for any level of VGS. Conversely, by using basic algebra we can obtain [from Eq. (5.3)] an equation for the resulting level of VGS for a given level of ID. The derivation is quite straight forward
and will result in
I
ID
VGS VP 1 (5.6)
DSS
Let us test Eq. (5.6) by finding the level of VGS that will result in a drain current
of 4.5 mA for the device with the characteristics of Fig. 5.15.
8m
A 4.5 mA
VGS 4 V 1 4 V(1 0
.5
625) 4 V(1 0.75)
4 V(0.25)
1 V
as substituted in the above calculation and verified by Fig. 5.15.
Shorthand Method
Since the transfer curve must be plotted so frequently, it would be quite advantageous
to have a shorthand method for plotting the curve in the quickest, most efficient manner while maintaining an acceptable degree of accuracy. The format of Eq. (5.3) is
such that specific levels of VGS will result in levels of ID that can be memorized to
provide the plot points needed to sketch the transfer curve. If we specify VGS to be
one-half the pinch-off value VP, the resulting level of ID will be the following, as determined by Shockley’s equation:
1 V /2
V
V
ID IDSS 1 GS
VP
IDSS
P
P
2
2
1
IDSS 1 2
2
IDSS(0.5)2
IDSS (0.25)
5.3 Transfer Characteristics
221
and
I SS
ID D
4
(5.7)
VGS VP/2
Now it is important to realize that Eq. (5.7) is not for a particular level of VP. It
is a general equation for any level of VP as long as VGS VP/2. The result specifies
that the drain current will always be one-fourth of the saturation level IDSS as long as
the gate-to-source voltage is one-half the pinch-off value. Note the level of ID for
VGS VP/2 4 V/2 2 V in Fig. 5.15.
If we choose ID IDSS/2 and substitute into Eq. (5.6), we find that
I
ID
VGS VP 1 D SS
I V (1 0.5) V (0.293)
VP 1 IDSS/2
P
P
D SS
and
VGS
0.3VP|ID IDSS /2
(5.8)
Additional points can be determined, but the transfer curve can be sketched to a
satisfactory level of accuracy simply using the four plot points defined above and reviewed in Table 5.1. In fact, in the analysis of Chapter 6, a maximum of four plot
points are used to sketch the transfer curves. On most occasions using just the plot
point defined by VGS VP /2 and the axis intersections at IDSS and VP will provide a
curve accurate enough for most calculations.
TABLE 5.1 VGS versus ID Using
Shockley’s Equation
EXAMPLE 5.1
VGS
ID
0
0.3 VP
0.5 VP
VP
IDSS
IDSS/2
IDSS/4
0 mA
Sketch the transfer curve defined by IDSS 12 mA and VP 6 V.
Solution
Two plot points are defined by
IDSS 12 mA
and
ID 0 mA
and
and
VGS 0 V
VGS VP
At VGS VP /2 6 V/2 3 V the drain current will be determined by ID IDSS /4 12 mA/4 3 mA. At ID IDSS /2 12 mA/2 6 mA the gate-to-source
voltage is determined by VGS 0.3VP 0.3(6 V) 1.8 V. All four plot points
are well defined on Fig. 5.16 with the complete transfer curve.
222
Chapter 5
Field-Effect Transistors
Figure 5.16 Transfer curve for
Example 5.1.
For p-channel devices Shockley’s equation (5.3) can still be applied exactly as it
appears. In this case, both VP and VGS will be positive and the curve will be the mirror
image of the transfer curve obtained with an n-channel and the same limiting values.
Sketch the transfer curve for a p-channel device with IDSS 4 mA and VP 3 V.
EXAMPLE 5.2
Solution
At VGS VP /2 3 V/2 1.5 V, ID IDSS /4 4 mA/4 1 mA. At ID IDSS/2 4 mA/2 2 mA, VGS 0.3VP 0.3(3 V) 0.9 V. Both plot points appear in Fig.
5.17 along with the points defined by IDSS and VP.
Figure 5.17 Transfer curve for the
p-channel device of Example 5.2.
5.4 SPECIFICATION SHEETS ( JFETs)
Although the general content of specification sheets may vary from the absolute
minimum to an extensive display of graphs and charts, there are a few fundamental
parameters that will be provided by all manufacturers. A few of the most important
are discussed in the following paragraphs. The specification sheet for the 2N5457
n-channel JFET as provided by Motorola is provided as Fig. 5.18.
5.4
Specification Sheets (JFETs)
223
Figure 5.18 2N5457 Motorola n-channel JFET.
Maximum Ratings
The maximum rating list usually appears at the beginning of the specification sheet,
with the maximum voltages between specific terminals, maximum current levels, and
the maximum power dissipation level of the device. The specified maximum levels
for VDS and VDG must not be exceeded at any point in the design operation of the device. The applied source VDD can exceed these levels, but the actual level of voltage
between these terminals must never exceed the level specified. Any good design will
224
Chapter 5
Field-Effect Transistors
try to avoid these levels by a good margin of safety. The term reverse in VGSR defines
the maximum voltage with the source positive with respect to the gate (as normally
biased for an n-channel device) before breakdown will occur. On some specification
sheets it is referred to as BVDSS —the Breakdown Voltage with the Drain-Source
Shorted (VDS 0 V). Although normally designed to operate with IG 0 mA, if forced
to accept a gate current it could withstand 10 mA before damage would occur. The
total device dissipation at 25°C (room temperature) is the maximum power the device can dissipate under normal operating conditions and is defined by
PD VDSID
(5.9)
Note the similarity in format with the maximum power dissipation equation for the
BJT transistor.
The derating factor is discussed in detail in Chapter 3, but for the moment recognize that the 2.82 mW/°C rating reveals that the dissipation rating decreases by
2.82 mW for each increase in temperature of 1°C above 25°C.
Electrical Characteristics
The electrical characteristics include the level of VP in the OFF CHARACTERISTICS and IDSS in the ON CHARACTERISTICS. In this case VP VGS(off) has a range
from 0.5 to 6.0 V and IDSS from 1 to 5 mA. The fact that both will vary from
device to device with the same nameplate identification must be considered in the
design process. The other quantities are defined under conditions appearing in parentheses. The small-signal characteristics are discussed in Chapter 9.
Case Construction and Terminal Identification
This particular JFET has the appearance provided on the specification sheet of Fig.
5.18. The terminal identification is also provided directly under the figure. JFETs are
also available in top-hat containers, as shown in Fig. 5.19 with its terminal identification.
Operating Region
The specification sheet and the curve defined by the pinch-off levels at each level of
VGS define the region of operation for linear amplification on the drain characteristics as shown in Fig. 5.20. The ohmic region defines the minimum permissible values of VDS at each level of VGS, and VDSmax specifies the maximum value for this pa-
Figure 5.19 Top-hat container
and terminal identification for a
p-channel JFET.
Figure 5.20 Normal operating
region for linear amplifier design.
5.4
Specification Sheets (JFETs)
225
rameter. The saturation current IDSS is the maximum drain current, and the maximum
power dissipation level defines the curve drawn in the same manner as described for
BJT transistors. The resulting shaded region is the normal operating region for amplifier design.
5.5 INSTRUMENTATION
Recall from Chapter 3 that hand-held instruments are available to measure the level
of dc for the BJT transistor. Similar instrumentation is not available to measure the
levels of IDSS and VP. However, the curve tracer introduced for the BJT transistor can
also display the drain characteristics of the JFET transistor through a proper setting
of the various controls. The vertical scale (in milliamperes) and the horizontal scale
(in volts) have been set to provide a full display of the characteristics, as shown in
Fig. 5.21. For the JFET of Fig. 5.21, each vertical division (in centimeters) reflects a
1-mA change in IC while each horizontal division has a value of 1 V. The step voltage is 500 mV/step (0.5 V/step), revealing that the top curve is defined by VGS 0 V
and the next curve down 0.5 V for the n-channel device. Using the same step voltage the next curve is 1 V, then 1.5 V, and finally 2 V. By drawing a line from
the top curve over to the ID axis, the level of IDSS can be estimated to be about 9 mA.
The level of VP can be estimated by noting the VGS value of the bottom curve and
taking into account the shrinking distance between curves as VGS becomes more and
more negative. In this case, VP is certainly more negative than 2 V and perhaps VP
is close to 2.5 V. However, keep in mind that the VGS curves contract very quickly
as they approach the cutoff condition, and perhaps VP 3 V is a better choice. It
VGS = 0 V
I DSS ≅ 9 mA
Vertical
Sens.
1 mA
per div.
VGS = –0.5 V
Horizontal
Sens.
1V
per div.
VGS = –1 V
I DSS = 4.5 mA
2
(VGS = – 0.9 V)
500 mV
per step.
VGS = –1.5 V
gm
2m
per div.
1 mA
div
VGS = –2 V
VP ≅ − 3 V
1V
div
Figure 5.21 Drain characteristics for a 2N4416 JFET transistor as displayed on a
curve tracer.
226
Chapter 5
Field-Effect Transistors
should also be noted that the step control is set for a 5-step display, limiting the displayed curves to VGS 0, 0.5, 1, 1.5, and 2 V. If the step control had been
increased to 10, the voltage per step could be reduced to 250 mV 0.25 V and the
curve for VGS 2.25 V would have been included as well as an additional curve
between each step of Fig. 5.21. The VGS 2.25 V curve would reveal how quickly
the curves are closing in on each other for the same step voltage. Fortunately, the level
of VP can be estimated to a reasonable degree of accuracy simply by applying a condition appearing in Table 5.1. That is, when ID IDSS /2, then VGS 0.3VP. For the
characteristics of Fig. 5.21, ID IDSS /2 9 mA/2 4.5 mA, and as visible from Fig.
5.21 the corresponding level of VGS is about 0.9 V. Using this information we find
that VP VGS /0.3 0.9 V/0.3 3 V, which will be our choice for this device.
Using this value we find that at VGS 2 V,
V
ID IDSS 1 GS
VP
2
2 V
9 mA 1 3 V
1 mA
2
as supported by Fig. 5.21.
At VGS 2.5 V, Shockley’s equation will result in ID 0.25 mA, with VP 3 V clearly revealing how quickly the curves contract near VP. The importance of
the parameter gm and how it is determined from the characteristics of Fig. 5.21 are
described in Chapter 8 when small-signal ac conditions are examined.
5.6 IMPORTANT RELATIONSHIPS
A number of important equations and operating characteristics have been introduced in
the last few sections that are of particular importance for the analysis to follow for the dc
and ac configurations. In an effort to isolate and emphasize their importance, they are repeated below next to a corresponding equation for the BJT transistor. The JFET equations
are defined for the configuration of Fig. 5.22a, while the BJT equations relate to Fig. 5.22b.
JFET
D
BJT
C
IC
ID
IG = 0 A
G
(
ID = IDSS 1 –
+
VGS
VGS
VP
IB
2
(
B
IS
IC = β IB
+
VBE = 0.7 V
–
IE
–
E
S
Figure 5.22 (a) JFET versus
(b) BJT.
(b)
(a)
JFET
V
1 GS
VP
BJT
2
⇔
IC IB
ID IS
⇔
IC
IG
⇔
VBE
ID IDSS
0A
IE
(5.10)
0.7 V
5.6 Important Relationships
227
A clear understanding of the impact of each of the equations above is sufficient
background to approach the most complex of dc configurations. Recall that VBE 0.7 V was often the key to initiating an analysis of a BJT configuration. Similarly,
the condition IG 0 A is often the starting point for the analysis of a JFET configuration. For the BJT configuration, IB is normally the first parameter to be determined.
For the JFET, it is normally VGS. The number of similarities between the analysis of
BJT and JFET dc configurations will become quite apparent in Chapter 6.
5.7 DEPLETION-TYPE MOSFET
As noted in the chapter introduction, there are two types of FETs: JFETs and MOSFETs. MOSFETs are further broken down into depletion type and enhancement type.
The terms depletion and enhancement define their basic mode of operation, while the
label MOSFET stands for metal-oxide-semiconductor-field-effect transistor. Since
there are differences in the characteristics and operation of each type of MOSFET,
they are covered in separate sections. In this section we examine the depletion-type
MOSFET, which happens to have characteristics similar to those of a JFET between
cutoff and saturation at IDSS but then has the added feature of characteristics that extend into the region of opposite polarity for VGS.
Basic Construction
The basic construction of the n-channel depletion-type MOSFET is provided in Fig.
5.23. A slab of p-type material is formed from a silicon base and is referred to as the
substrate. It is the foundation upon which the device will be constructed. In some
cases the substrate is internally connected to the source terminal. However, many discrete devices provide an additional terminal labeled SS, resulting in a four-terminal
device, such as that appearing in Fig. 5.23. The source and drain terminals are connected through metallic contacts to n-doped regions linked by an n-channel as shown
in the figure. The gate is also connected to a metal contact surface but remains insulated from the n-channel by a very thin silicon dioxide (SiO2) layer. SiO2 is a particular type of insulator referred to as a dielectric that sets up opposing (as revealed by
Figure 5.23 n-Channel depletion-type MOSFET.
228
Chapter 5
Field-Effect Transistors
the prefix di-) electric fields within the dielectric when exposed to an externally applied field. The fact that the SiO2 layer is an insulating layer reveals the following
fact:
There is no direct electrical connection between the gate terminal and the
channel of a MOSFET.
In addition:
It is the insulating layer of SiO2 in the MOSFET construction that accounts
for the very desirable high input impedance of the device.
In fact, the input resistance of a MOSFET is often that of the typical JFET, even
though the input impedance of most JFETs is sufficiently high for most applications.
The very high input impedance continues to fully support the fact that the gate current (IG) is essentially zero amperes for dc-biased configurations.
The reason for the label metal-oxide-semiconductor FET is now fairly obvious:
metal for the drain, source, and gate connections to the proper surface—in particular, the gate terminal and the control to be offered by the surface area of the contact,
the oxide for the silicon dioxide insulating layer, and the semiconductor for the basic
structure on which the n- and p-type regions are diffused. The insulating layer between the gate and channel has resulted in another name for the device: insulatedgate FET or IGFET, although this label is used less and less in current literature.
Basic Operation and Characteristics
In Fig. 5.24 the gate-to-source voltage is set to zero volts by the direct connection
from one terminal to the other, and a voltage VDS is applied across the drain-to-source
terminals. The result is an attraction for the positive potential at the drain by the free
electrons of the n-channel and a current similar to that established through the channel of the JFET. In fact, the resulting current with VGS 0 V continues to be labeled
IDSS, as shown in Fig. 5.25.
Figure 5.24 n-Channel depletion-type MOSFET with VGS 0 V and an applied
voltage VDD.
5.7 Depletion-Type MOSFET
229
ID (mA)
Depletion
mode
ID
VGS = + 1 V
10.9
Enhancement
mode
I DSS
8
VGS = 0 V
VGS = –1 V
I DSS
2
IDSS
4
4
2
– 6 – 5 – 4 – 3 – 2 –1 0
VP 0.3V
VP
P
2
1
VGS
VGS = – 2 V
VGS = VP = – 3 V
2
–4 V
–5 V
0
VGS = VP = – 6 V
VDS
Figure 5.25 Drain and transfer characteristics for an n-channel depletion-type MOSFET.
In Fig. 5.26, VGS has been set at a negative voltage such as 1 V. The negative
potential at the gate will tend to pressure electrons toward the p-type substrate (like
charges repel) and attract holes from the p-type substrate (opposite charges attract) as
shown in Fig. 5.26. Depending on the magnitude of the negative bias established by
VGS, a level of recombination between electrons and holes will occur that will reduce
the number of free electrons in the n-channel available for conduction. The more negative the bias, the higher the rate of recombination. The resulting level of drain current is therefore reduced with increasing negative bias for VGS as shown in Fig. 5.25
for VGS 1 V, 2 V, and so on, to the pinch-off level of 6 V. The resulting levels of drain current and the plotting of the transfer curve proceeds exactly as described
for the JFET.
Figure 5.26 Reduction in free
carriers in channel due to a negative potential at the gate terminal.
For positive values of VGS , the positive gate will draw additional electrons (free
carriers) from the p-type substrate due to the reverse leakage current and establish
new carriers through the collisions resulting between accelerating particles. As the
gate-to-source voltage continues to increase in the positive direction, Fig. 5.25 reveals
that the drain current will increase at a rapid rate for the reasons listed above. The
230
Chapter 5
Field-Effect Transistors
vertical spacing between the VGS 0 V and VGS 1 V curves of Fig. 5.25 is a clear
indication of how much the current has increased for the 1-V change in VGS. Due
to the rapid rise, the user must be aware of the maximum drain current rating since
it could be exceeded with a positive gate voltage. That is, for the device of Fig. 5.25,
the application of a voltage VGS 4 V would result in a drain current of 22.2 mA,
which could possibly exceed the maximum rating (current or power) for the device.
As revealed above, the application of a positive gate-to-source voltage has “enhanced”
the level of free carriers in the channel compared to that encountered with VGS 0 V. For this reason the region of positive gate voltages on the drain or transfer characteristics is often referred to as the enhancement region, with the region between
cutoff and the saturation level of IDSS referred to as the depletion region.
It is particularly interesting and helpful that Shockley’s equation will continue to
be applicable for the depletion-type MOSFET characteristics in both the depletion
and enhancement regions. For both regions, it is simply necessary that the proper sign
be included with VGS in the equation and the sign be carefully monitored in the mathematical operations.
Sketch the transfer characteristics for an n-channel depletion-type MOSFET with
IDSS 10 mA and VP 4 V.
EXAMPLE 5.3
Solution
At VGS 0 V,
ID IDSS 10 mA
VGS VP 4 V,
ID 0 mA
VP
4 V
VGS 2 V,
2
2
I SS
and at ID D,
2
I SS
10 mA
ID D
2.5 mA
4
4
VGS 0.3VP 0.3(4 V) 1.2 V
all of which appear in Fig. 5.27.
Before plotting the positive region of VGS, keep in mind that ID increases very
rapidly with increasing positive values of VGS. In other words, be conservative with
the choice of values to be substituted into Shockley’s equation. In this case, we will
try 1 V as follows:
V
ID IDSS 1 GS
VP
2
1 V
10 mA 1 4 V
2
10 mA(1 0.25)2 10 mA(1.5625)
15.63 mA
which is sufficiently high to finish the plot.
p-Channel Depletion-Type MOSFET
The construction of a p-channel depletion-type MOSFET is exactly the reverse of that
appearing in Fig. 5.23. That is, there is now an n-type substrate and a p-type channel, as shown in Fig. 5.28a. The terminals remain as identified, but all the voltage polarities and the current directions are reversed, as shown in the same figure. The drain
characteristics would appear exactly as in Fig. 5.25 but with VDS having negative val5.7 Depletion-Type MOSFET
Figure 5.27 Transfer characteristics for an n-channel depletiontype MOSFET with IDSS 10 mA
and VP 4 V.
231
ID (mA)
ID (mA)
D
VGS
8
8
6
p
+
9
n
p
p
SS
IDSS
VGS = 0 V
6
5
5
4
4
3
3
2
2
1
1
–1 0
–
VGS = –1 V
7
7
ID
G
9
1 2 3 4 5 6
VP
VGS
VGS = +1 V
VGS = +2 V
VGS = +3 V
VGS = +4 V
VGS = +5 V
0
VGS = VP = +6 V
VDS
S
(a)
(b)
(c)
Figure 5.28 p-Channel depletion-type MOSFET with IDSS 6 mA and VP 6 V.
ues, ID having positive values as indicated (since the defined direction is now reversed), and VGS having the opposite polarities as shown in Fig. 5.28c. The reversal
in VGS will result in a mirror image (about the ID axis) for the transfer characteristics
as shown in Fig. 5.28b. In other words, the drain current will increase from cutoff at
VGS VP in the positive VGS region to IDSS and then continue to increase for increasingly negative values of VGS. Shockley’s equation is still applicable and requires
simply placing the correct sign for both VGS and VP in the equation.
Symbols, Specification Sheets, and Case
Construction
The graphic symbols for an n- and p-channel depletion-type MOSFET are provided
in Fig. 5.29. Note how the symbols chosen try to reflect the actual construction of the
device. The lack of a direct connection (due to the gate insulation) between the gate
and channel is represented by a space between the gate and the other terminals of the
symbol. The vertical line representing the channel is connected between the drain and
source and is “supported” by the substrate. Two symbols are provided for each type
of channel to reflect the fact that in some cases the substrate is externally available
while in others it is not. For most of the analysis to follow in Chapter 6, the substrate
and source will be connected and the lower symbols will be employed.
Figure 5.29 Graphic symbols
for (a) n-channel depletion-type
MOSFETs and (b) p-channel
depletion-type MOSFETs.
232
Chapter 5
Field-Effect Transistors
The device appearing in Fig. 5.30 has three terminals, with the terminal identification appearing in the same figure. The specification sheet for a depletion-type MOSFET is similar to that of a JFET. The levels of VP and IDSS are provided along with
a list of maximum values and typical “on” and “off” characteristics. In addition, how-
Figure 5.30 2N3797 Motorola n-channel depletion-type MOSFET.
5.7 Depletion-Type MOSFET
233
ever, since ID can extend beyond the IDSS level, another point is normally provided
that reflects a typical value of ID for some positive voltage (for an n-channel device).
For the unit of Fig. 5.30, ID is specified as ID(on) 9 mA dc, with VDS 10 V and
VGS 3.5 V.
5.8 ENHANCEMENT-TYPE MOSFET
Although there are some similarities in construction and mode of operation between
depletion-type and enhancement-type MOSFETs, the characteristics of the enhancement-type MOSFET are quite different from anything obtained thus far. The transfer
curve is not defined by Shockley’s equation, and the drain current is now cut off until the gate-to-source voltage reaches a specific magnitude. In particular, current control in an n-channel device is now effected by a positive gate-to-source voltage rather
than the range of negative voltages encountered for n-channel JFETs and n-channel
depletion-type MOSFETs.
Basic Construction
The basic construction of the n-channel enhancement-type MOSFET is provided in
Fig. 5.31. A slab of p-type material is formed from a silicon base and is again referred to as the substrate. As with the depletion-type MOSFET, the substrate is sometimes internally connected to the source terminal, while in other cases a fourth lead
is made available for external control of its potential level. The source and drain terminals are again connected through metallic contacts to n-doped regions, but note in
Fig. 5.31 the absence of a channel between the two n-doped regions. This is the primary difference between the construction of depletion-type and enhancement-type
MOSFETs—the absence of a channel as a constructed component of the device. The
SiO2 layer is still present to isolate the gate metallic platform from the region between the drain and source, but now it is simply separated from a section of the
p-type material. In summary, therefore, the construction of an enhancement-type
MOSFET is quite similar to that of the depletion-type MOSFET, except for the absence of a channel between the drain and source terminals.
Figure 5.31 n-Channel enhancement-type MOSFET.
234
Chapter 5
Field-Effect Transistors
Basic Operation and Characteristics
If VGS is set at 0 V and a voltage applied between the drain and source of the device
of Fig. 5.31, the absence of an n-channel (with its generous number of free carriers)
will result in a current of effectively zero amperes—quite different from the depletion-type MOSFET and JFET where ID IDSS. It is not sufficient to have a large accumulation of carriers (electrons) at the drain and source (due to the n-doped regions)
if a path fails to exist between the two. With VDS some positive voltage, VGS at 0 V,
and terminal SS directly connected to the source, there are in fact two reverse-biased
p-n junctions between the n-doped regions and the p-substrate to oppose any significant flow between drain and source.
In Fig. 5.32 both VDS and VGS have been set at some positive voltage greater than
0 V, establishing the drain and gate at a positive potential with respect to the source.
The positive potential at the gate will pressure the holes (since like charges repel) in
the p-substrate along the edge of the SiO2 layer to leave the area and enter deeper regions of the p-substrate, as shown in the figure. The result is a depletion region near
the SiO2 insulating layer void of holes. However, the electrons in the p-substrate (the
minority carriers of the material) will be attracted to the positive gate and accumulate in the region near the surface of the SiO2 layer. The SiO2 layer and its insulating qualities will prevent the negative carriers from being absorbed at the gate terminal. As VGS increases in magnitude, the concentration of electrons near the SiO2 surface
increases until eventually the induced n-type region can support a measurable flow
between drain and source. The level of VGS that results in the significant increase in
drain current is called the threshold voltage and is given the symbol VT. On specification sheets it is referred to as VGS(Th), although VT is less unwieldy and will be used
in the analysis to follow. Since the channel is nonexistent with VGS 0 V and “enhanced” by the application of a positive gate-to-source voltage, this type of MOSFET
is called an enhancement-type MOSFET. Both depletion- and enhancement-type MOSFETs have enhancement-type regions, but the label was applied to the latter since it
is its only mode of operation.
Electrons attracted to positive gate
(induced n-channel)
Region depleted of p-type
carriers (holes)
ID
n
D
IG = 0 A
G
+
VGS
–
S
+
+
+
+
+
+
e
e
e
e
e
e
e
e
+
+
+
+
+
+
+
SS
p
+
VDS
–
n
IS = ID
Insulating layer
Holes repelled
by positive gate
Figure 5.32 Channel formation
in the n-channel enhancementtype MOSFET.
5.8 Enhancement-Type MOSFET
235
As VGS is increased beyond the threshold level, the density of free carriers in the
induced channel will increase, resulting in an increased level of drain current. However, if we hold VGS constant and increase the level of VDS, the drain current will
eventually reach a saturation level as occurred for the JFET and depletion-type MOSFET. The leveling off of ID is due to a pinching-off process depicted by the narrower
channel at the drain end of the induced channel as shown in Fig. 5.33. Applying Kirchhoff’s voltage law to the terminal voltages of the MOSFET of Fig. 5.33, we find that
VDG VDS VGS
(5.11)
Figure 5.33 Change in channel
and depletion region with increasing level of VDS for a fixed value
of VGS.
If VGS is held fixed at some value such as 8 V and VDS is increased from 2 to 5
V, the voltage VDG [by Eq. (5.11)] will drop from 6 to 3 V and the gate will become less and less positive with respect to the drain. This reduction in gate-to-drain
voltage will in turn reduce the attractive forces for free carriers (electrons) in this region of the induced channel, causing a reduction in the effective channel width. Eventually, the channel will be reduced to the point of pinch-off and a saturation condition will be established as described earlier for the JFET and depletion-type MOSFET.
In other words, any further increase in VDS at the fixed value of VGS will not affect
the saturation level of ID until breakdown conditions are encountered.
The drain characteristics of Fig. 5.34 reveal that for the device of Fig. 5.33 with
VGS 8 V, saturation occurred at a level of VDS 6 V. In fact, the saturation level
for VDS is related to the level of applied VGS by
VDSsat VGS VT
(5.12)
Obviously, therefore, for a fixed value of VT, then the higher the level of VGS, the more
the saturation level for VDS, as shown in Fig. 5.33 by the locus of saturation levels.
236
Chapter 5
Field-Effect Transistors
ID (mA)
Locus of VDSsat
11
VGS = +8 V
10
9
8
VGS = +7 V
7
6
5
VGS = +6 V
4
3
VGS = +5 V
2
VGS = +4 V
VGS = +3 V
1
0
5V
6V
10 V
15 V
20 V
25 V
VGS = V T = 2 V
VDS
Figure 5.34 Drain characteristics of an n-channel enhancement-type
MOSFET with VT 2 V and k 0.278 103 A/V2.
For the characteristics of Fig. 5.33 the level of VT is 2 V, as revealed by the fact
that the drain current has dropped to 0 mA. In general, therefore:
For values of VGS less than the threshold level, the drain current of an enhancement-type MOSFET is 0 mA.
Figure 5.34 clearly reveals that as the level of VGS increased from VT to 8 V, the
resulting saturation level for ID also increased from a level of 0 to 10 mA. In addition, it is quite noticeable that the spacing between the levels of VGS increased as the
magnitude of VGS increased, resulting in ever-increasing increments in drain current.
For levels of VGS VT, the drain current is related to the applied gate-to-source
voltage by the following nonlinear relationship:
ID k(VGS VT)2
(5.13)
Again, it is the squared term that results in the nonlinear (curved) relationship between ID and VGS. The k term is a constant that is a function of the construction of
the device. The value of k can be determined from the following equation [derived
from Eq. (5.13)] where ID(on) and VGS(on) are the values for each at a particular point
on the characteristics of the device.
ID(on)
k 2
(VGS(on) VT)
(5.14)
Substituting ID(on) 10 mA when VGS(on) 8 V from the characteristics of Fig.
5.34 yields
10 mA
10 mA
10 mA
k 2 (8 V 2 V)
(6 V)2
36 V2
0.278 103 A/V2
and a general equation for ID for the characteristics of Fig. 5.34 results in:
ID 0.278
103(VGS 2 V)2
5.8 Enhancement-Type MOSFET
237
Substituting VGS 4 V, we find that
103(4 V 2 V)2 0.278
ID 0.278
103(2)2
103(4) 1.11 mA
0.278
as verified by Fig. 5.34. At VGS VT , the squared term is 0 and ID 0 mA.
For the dc analysis of enhancement-type MOSFETs to appear in Chapter 6, the
transfer characteristics will again be the characteristics to be employed in the graphical solution. In Fig. 5.35 the drain and transfer characteristics have been set side by
side to describe the transfer process from one to the other. Essentially, it proceeds as
introduced earlier for the JFET and depletion-type MOSFETs. In this case, however,
it must be remembered that the drain current is 0 mA for VGS VT. At this point a
measurable current will result for ID and will increase as defined by Eq. (5.13). Note
that in defining the points on the transfer characteristics from the drain characteristics, only the saturation levels are employed, thereby limiting the region of operation
to levels of VDS greater than the saturation levels as defined by Eq. (5.12).
ID (mA)
ID (mA)
10
10
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
1
2
VT
3
4
5
6
7
8
VGS
VGS = +8 V
VGS = +7 V
VGS = +6 V
VGS = +5 V
VGS = +4 V
VGS = +3 V
0
5
10
15
20
25
VDS
VGS = V T = 2 V
Figure 5.35 Sketching the transfer characteristics for an n-channel enhancementtype MOSFET from the drain characteristics.
The transfer curve of Fig. 5.35 is certainly quite different from those obtained earlier. For an n-channel (induced) device, it is now totally in the positive VGS region
and does not rise until VGS VT. The question now surfaces as to how to plot the
transfer characteristics given the levels of k and VT as included below for a particular MOSFET:
ID 0.5
103(VGS 4 V)2
First, a horizontal line is drawn at ID 0 mA from VGS 0 V to VGS 4 V as
shown in Fig. 5.36a. Next, a level of VGS greater than VT such as 5 V is chosen and
substituted into Eq. (5.13) to determine the resulting level of ID as follows:
ID 0.5
0.5
103(VGS 4 V)2
103(5 V 4 V)2 0.5
0.5 mA
238
Chapter 5
Field-Effect Transistors
103(1)2
Figure 5.36 Plotting the transfer characteristics of an n-channel enhancementtype MOSFET with k 0.5 103 A/V2 and VT 4 V.
and a point on the plot is obtained as shown in Fig. 5.36b. Finally, additional levels
of VGS are chosen and the resulting levels of ID obtained. In particular, at VGS 6,
7, and 8 V, the level of ID is 2, 4.5, and 8 mA, respectively, as shown on the resulting plot of Fig. 5.36c.
p-Channel Enhancement-Type MOSFETs
The construction of a p-channel enhancement-type MOSFET is exactly the reverse of
that appearing in Fig. 5.31, as shown in Fig. 5.37a. That is, there is now an n-type
substrate and p-doped regions under the drain and source connections. The terminals
remain as identified, but all the voltage polarities and the current directions are reversed. The drain characteristics will appear as shown in Fig. 5.37c, with increasing
levels of current resulting from increasingly negative values of VGS. The transfer characteristics will be the mirror image (about the ID axis) of the transfer curve of Fig.
5.35, with ID increasing with increasingly negative values of VGS beyond VT, as shown
in Fig. 5.37b. Equations (5.11) through (5.14) are equally applicable to p-channel devices.
5.8 Enhancement-Type MOSFET
239
ID (mA)
D
ID
p
G
n
–
SS
p
+
–6 –5 –4 –3 –2 –1
VT
D
(a)
ID (mA)
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
VGS
VGS = –6 V
VGS = –5 V
VGS = –4 V
VGS = –3 V
0
VGS = VT = –2 V
(b)
VDS
(c)
Figure 5.37 p-Channel enhancement-type MOSFET with VT 2 V and
k 0.5
103 A/V2.
Symbols, Specification Sheets, and Case
Construction
The graphic symbols for the n- and p-channel enhancement-type MOSFETs are provided as Fig. 5.38. Again note how the symbols try to reflect the actual construction
of the device. The dashed line between drain and source was chosen to reflect the fact
that a channel does not exist between the two under no-bias conditions. It is, in fact,
the only difference between the symbols for the depletion-type and enhancement-type
MOSFETs.
Figure 5.38 Symbols for (a)
n-channel enhancement-type
MOSFETs and (b) p-channel
enhancement-type MOSFETs.
The specification sheet for a Motorola n-channel enhancement-type MOSFET is
provided as Fig. 5.39. The case construction and terminal identification are provided
next to the maximum ratings, which now include a maximum drain current of 30 mA
dc. The specification sheet provides the level of IDSS under “off” conditions, which is
now simply 10 nA dc (at VDS 10 V and VGS 0 V) compared to the milliampere
range for the JFET and depletion-type MOSFET. The threshold voltage is specified
240
Chapter 5
Field-Effect Transistors
Figure 5.39 2N4351 Motorola n-channel enhancement-type MOSFET.
as VGS(Th) and has a range of 1 to 5 V dc, depending on the unit employed. Rather
than provide a range of k in Eq. (5.13), a typical level of ID(on) (3 mA in this case) is
specified at a particular level of VGS(on) (10 V for the specified ID level). In other
words, when VGS 10 V, ID 3 mA. The given levels of VGS(Th), ID(on), and VGS(on)
permit a determination of k from Eq. (5.14) and a writing of the general equation for
the transfer characteristics. The handling requirements of MOSFETs are reviewed in
Section 5.9.
5.8 Enhancement-Type MOSFET
241
EXAMPLE 5.4
Using the data provided on the specification sheet of Fig. 5.39 and an average threshold voltage of VGS(Th) 3 V, determine:
(a) The resulting value of k for the MOSFET.
(b) The transfer characteristics.
Solution
ID(on)
(a) Eq. (5.14): k (VGS(on) VGS(Th) )2
3 103
3 mA
3 mA
A/V2
(10 V 3 V)2
(7 V)2
49
0.061 103 A/V2
(b) Eq. (5.13): ID k(VGS VT)2
0.061 103(VGS 3 V)2
For VGS 5 V,
ID 0.061
0.061
103(5 V 3 V)2 0.061
103(2)2
103(4) 0.244 mA
For VGS 8, 10, 12, and 14 V, ID will be 1.525, 3 (as defined), 4.94, and 7.38 mA,
respectively. The transfer characteristics are sketched in Fig. 5.40.
Figure 5.40 Solution to Example 5.4.
5.9 MOSFET HANDLING
The thin SiO2 layer between the gate and channel of MOSFETs has the positive effect of providing a high-input-impedance characteristic for the device, but because of
its extremely thin layer, it introduces a concern for its handling that was not present
for the BJT or JFET transistors. There is often sufficient accumulation of static charge
(that we pick up from our surroundings) to establish a potential difference across the
thin layer that can break down the layer and establish conduction through it. It is
therefore imperative that we leave the shorting (or conduction) shipping foil (or ring)
242
Chapter 5
Field-Effect Transistors
connecting the leads of the device together until the device is to be inserted in the
system. The shorting ring prevents the possibility of applying a potential across any
two terminals of the device. With the ring the potential difference between any two
terminals is maintained at 0 V. At the very least always touch ground to permit discharge of the accumulated static charge before handling the device, and always pick
up the transistor by the casing.
There are often transients (sharp changes in voltage or current) in a network when
elements are removed or inserted if the power is on. The transient levels can often be
more than the device can handle, and therefore the power should always be off when
network changes are made.
The maximum gate-to-source voltage is normally provided in the list of maximum
ratings of the device. One method of ensuring that this voltage is not exceeded (perhaps by transient effects) for either polarity is to introduce two Zener diodes, as shown
in Fig. 5.41. The Zeners are back to back to ensure protection for either polarity. If
both are 30-V Zeners and a positive transient of 40 V appears, the lower Zener will
“fire” at 30 V and the upper will turn on with a 0-V drop (ideally—for the positive
“on” region of a semiconductor diode) across the other diode. The result is a maximum of 30 V for the gate-to-source voltage. One disadvantage introduced by the Zener
protection is that the off resistance of a Zener diode is less than the input impedance
established by the SiO2 layer. The result is a reduction in input resistance, but even
so it is still high enough for most applications. So many of the discrete devices now
have the Zener protection that some of the concerns listed above are not as troublesome. However, it is still best to be somewhat cautious when handling discrete MOSFET devices.
Figure 5.41 Zener-protected
MOSFET.
5.10 VMOS
One of the disadvantages of the typical MOSFET is the reduced power-handling levels (typically, less than 1 W) compared to BJT transistors. This shortfall for a device
with so many positive characteristics can be softened by changing the construction
mode from one of a planar nature such as shown in Fig. 5.23 to one with a vertical
structure as shown in Fig. 5.42. All the elements of the planar MOSFET are present
in the vertical metal-oxide-silicon FET (VMOS)—the metallic surface connection to
the terminals of the device—the SiO2 layer between the gate and the p-type region
between the drain and source for the growth of the induced n-channel (enhancement-
Figure 5.42 VMOS construction.
5.10 VMOS
243
mode operation). The term vertical is due primarily to the fact that the channel is now
formed in the vertical direction rather than the horizontal direction for the planar device. However, the channel of Fig. 5.42 also has the appearance of a “V” cut in the
semiconductor base, which often stands out as a characteristic for mental memorization of the name of the device. The construction of Fig. 5.42 is somewhat simplistic
in nature, leaving out some of the transition levels of doping, but it does permit a description of the most important facets of its operation.
The application of a positive voltage to the drain and a negative voltage to the
source with the gate at 0 V or some typical positive “on” level as shown in Fig. 5.42
will result in the induced n-channel in the narrow p-type region of the device. The
length of the channel is now defined by the vertical height of the p-region, which can
be made significantly less than that of a channel using planar construction. On a horizontal plane the length of the channel is limited to 1 to 2 m (1 m 106 m). Diffusion layers (such as the p-region of Fig. 5.42) can be controlled to small fractions
of a micrometer. Since decreasing channel lengths result in reduced resistance levels,
the power dissipation level of the device (power lost in the form of heat) at operating current levels will be reduced. In addition, the contact area between the channel
and the n region is greatly increased by the vertical mode construction, contributing to a further decrease in the resistance level and an increased area for current between the doping layers. There is also the existence of two conduction paths between
drain and source, as shown in Fig. 5.42, to further contribute to a higher current rating. The net result is a device with drain currents that can reach the ampere levels
with power levels exceeding 10 W.
In general:
Compared with commercially available planar MOSFETs, VMOS FETs have
reduced channel resistance levels and higher current and power ratings.
An additional important characteristic of the vertical construction is:
VMOS FETs have a positive temperature coefficient that will combat the possibility of thermal runaway.
If the temperature of a device should increase due to the surrounding medium or
currents of the device, the resistance levels will increase, causing a reduction in drain
current rather than an increase as encountered for a conventional device. Negative
temperature coefficients result in decreased levels of resistance with increases in temperature that fuel the growing current levels and result in further temperature instability and thermal runaway.
Another positive characteristic of the VMOS configuration is:
The reduced charge storage levels result in faster switching times for VMOS
construction compared to those for conventional planar construction.
In fact, VMOS devices typically have switching times less than one-half that encountered for the typical BJT transistor.
5.11 CMOS
A very effective logic circuit can be established by constructing a p-channel and an
n-channel MOSFET on the same substrate as shown in Fig. 5.43. Note the induced
p-channel on the left and the induced n-channel on the right for the p- and n-channel
devices, respectively. The configuration is referred to as a complementary MOSFET
arrangement (CMOS) that has extensive applications in computer logic design. The
relatively high input impedance, fast switching speeds, and lower operating power
levels of the CMOS configuration have resulted in a whole new discipline referred to
as CMOS logic design.
244
Chapter 5
Field-Effect Transistors
Vi
Vo
VSS
G2
S2
G1
D2
D1
S1
SiO2
n+
p
p+
p+
n
n+
When "on"
n+
When "on"
p-channel MOSFET
p+
p
n-channel MOSFET
n-type substrate
Figure 5.43 CMOS with the connections indicated in Fig. 5.44.
One very effective use of the complementary arrangement is as an inverter, as
shown in Fig. 5.44. As introduced for switching transistors, an inverter is a logic element that “inverts” the applied signal. That is, if the logic levels of operation are
0 V (0-state) and 5 V (1-state), an input level of 0 V will result in an output level of
5 V, and vice versa. Note in Fig. 5.44 that both gates are connected to the applied signal and both drain to the output Vo. The source of the p-channel MOSFET (Q2) is
connected directly to the applied voltage VSS, while the source of the n-channel MOSFET (Q1) is connected to ground. For the logic levels defined above, the application
of 5 V at the input should result in approximately 0 V at the output. With 5 V at Vi
(with respect to ground), VGS1 Vi and Q1 is “on,” resulting in a relatively low resistance between drain and source as shown in Fig. 5.45. Since Vi and VSS are at 5 V,
VGS2 0 V, which is less than the required VT for the device, resulting in an “off”
state. The resulting resistance level between drain and source is quite high for Q2, as
shown in Fig. 5.45. A simple application of the voltage-divider rule will reveal that
Vo is very close to 0 V or the 0-state, establishing the desired inversion process. For
an applied voltage Vi of 0 V (0-state), VGS1 0 V and Q1 will be off with VSS2 5 V, turning on the p-channel MOSFET. The result is that Q2 will present a small
resistance level, Q1 a high resistance, and Vo VSS 5 V (the 1-state). Since the
drain current that flows for either case is limited by the “off” transistor to the leakage value, the power dissipated by the device in either state is very low. Additional
comment on the application of CMOS logic is presented in Chapter 17.
VSS
Q2 off
5V
Ileakage
R2 (high)
Vo =
Q1 on
Figure 5.44 CMOS inverter.
R1VSS
R1 + R2
≅ 0 V (0-state)
R1 (low)
Figure 5.45 Relative resistance
levels for Vi 5 V (1-state).
5.11 CMOS
245
5.12 SUMMARY TABLE
Since the transfer curves and some important characteristics vary from one type of
FET to another, Table 5.2 was developed to clearly display the differences from one
device to the next. A clear understanding of all the curves and parameters of the table
will provide a sufficient background for the dc and ac analyses to follow in Chapters
6 and 8. Take a moment to ensure that each curve is recognizable and its derivation
understood, and then establish a basis for comparison of the levels of the important
parameters of Ri and Ci for each device.
TABLE 5.2 Field Effect Transistors
Type
-SymbolBasic Relationships
Transfer Curve
Input Resistance
and Capacitance
JFET
(n-channel)
Ri 100 M
Ci: (1 10) pF
MOSFET
depletion-type
(n-channel)
Ri 1010 Ci: (1 10) pF
MOSFET
enhancement-type
(n-channel)
Ri 1010 Ci: (1 10) pF
246
Chapter 5
Field-Effect Transistors
5.13 PSPICE WINDOWS
The characteristics of an n-channel JFET can be found in much the same manner as
employed for the bipolar transistor. The series of curves for various levels of V will
require a nested sweep under the main sweep for the drain-to-source voltage. The
configuration required appears in Fig. 5.46. Note the absence of any resistors since
the input impedance is assumed to be infinite, resulting in IG 0 A. Calling up the
device specifications through Edit-Model-Edit Instance Model (Text) will result in
a display having at the head of the listing a parameter Beta. For the junction-fieldeffect transistor Beta is defined by
IDSS
Beta 2
Vp
(5.15)
The parameter Vto 3 defines VGS VP 3 V as the pinch-off voltage—
something to check when we obtain our characteristics. Choosing the Setup Analysis
icon (recall that it has the horizontal blue line at the top), the DC Sweep is first enabled
and then activated to produce the DC Sweep dialog box. Select Voltage SourceLinear, and insert the Name: VDD, the Start Value of 0 V, End Value of 10 V, and
Increment of 0.01 V. Then, the Nested Sweep is chosen, and Voltage and Linear are
chosen once more. Finally, the Name: VGG is entered, the Start Value of 0 V is chosen, the End Value of 5 V is entered, and the Increment is set at 1 V. Then, be
sure to Enable Nested Sweep before clicking on OK and closing. With the Automatically run Probe after Simulation enabled, clicking on the analysis icon will result in
the OrCAD-MicroSim Probe screen. There is no need to call up the X-Axis Settings
because the horizontal axis has the correct range and the voltage VDD is actually the
drain-to-source voltage. By choosing the Trace icon, the Add Traces dialog box will
appear. ID(J1) is chosen, followed by OK. The result is the set of characteristics appearing in Fig. 5.47. The remaining labels were added using the ABC icon.
Note that the pinch-off voltage is 3 V, as expected by the Vto parameter. The
value of IDSS is very close to 12 mA.
Figure 5.46 Network employed
to obtain the characteristics of the
n-channel J2N3819 JFET.
Figure 5.47 Drain characteristics for the n-channel J2N3819 JFET of Figure 5.46.
5.13 PSpice Windows
247
The transfer characteristics can be obtained by returning to the network configuration and choosing the Analysis-Setup icon. The DC Sweep is again enabled, and
the DC Sweep is chosen. This time, since the result will only be one curve, a nested
operation will not be performed. After choosing Voltage Source and Linear, the Name
will be VGG, the Start Value 3 V (since we now know that VP 3 V), the End
Value 0 V, and the Increment 0.01 V to get a good continuous plot. After an OK followed by a Close, the Simulation icon can be chosen. Once the Probe screen appears, choose Plot-X-Axis Settings-Axis Variable and choose V(J1:g) for the gateto-source voltage. Choose OK and we’re back to the X-Axis Settings dialog box to
choose the User Defined range of 3 V to 0 V (which already appears because of
our sweep settings). Choose OK again and the Trace ID(J1) can be chosen to result
in the transfer characteristics of Fig. 5.48.
Figure 5.48 Transfer characteristics for the n-channel J2N3819 JFET of Figure 5.46.
248
Chapter 5
Field-Effect Transistors
PROBLEMS
§ 5.2 Construction and Characteristics of JFETs
1. (a) Draw the basic construction of a p-channel JFET.
(b) Apply the proper biasing between drain and source and sketch the depletion region for
VGS 0 V.
2. Using the characteristics of Fig. 5.10, determine ID for the following levels of VGS (with
VDS VP).
(a) VGS 0 V.
(b) VGS 1 V.
(c) VGS 1.5 V.
(d) VGS 1.8 V.
(e) VGS 4 V.
(f) VGS 6 V.
3. (a) Determine VDS for VGS 0 V and ID 6 mA using the characteristics of Fig. 5.10.
(b) Using the results of part (a), calculate the resistance of the JFET for the region ID 0 to
6 mA for VGS 0 V.
(c) Determine VDS for VGS 1 V and ID 3 mA.
(d) Using the results of part (c), calculate the resistance of the JFET for the region ID 0 to
3 mA for VGS 1 V.
(e) Determine VDS for VGS 2 V and ID 1.5 mA.
(f) Using the results of part (e), calculate the resistance of the JFET for the region ID 0 to
1.5 mA for VGS 2 V.
(g) Defining the result of part (b) as ro, determine the resistance for VGS 1 V using Eq.
(5.1) and compare with the results of part (d).
(h) Repeat part (g) for VGS 2 V using the same equation, and compare the results with part
(f).
(i) Based on the results of parts (g) and (h), does Eq. (5.1) appear to be a valid approximation?
4. Using the characteristics of Fig. 5.10:
(a) Determine the difference in drain current (for VDS VP) between VGS 0 V and VGS 1 V.
(b) Repeat part (a) between VGS 1 and 2 V.
(c) Repeat part (a) between VGS 2 and 3 V.
(d) Repeat part (a) between VGS 3 and 4 V.
(e) Is there a marked change in the difference in current levels as VGS becomes increasingly
negative?
(f) Is the relationship between the change in VGS and the resulting change in ID linear or nonlinear? Explain.
5. What are the major differences between the collector characteristics of a BJT transistor and the
drain characteristics of a JFET transistor? Compare the units of each axis and the controlling
variable. How does IC react to increasing levels of IB versus changes in ID to increasingly negative values of VGS? How does the spacing between steps of IB compare to the spacing between
steps of VGS? Compare VCsat to VP in defining the nonlinear region at low levels of output voltage.
6. (a) Describe in your own words why IG is effectively zero amperes for a JFET transistor.
(b) Why is the input impedance to a JFET so high?
(c) Why is the terminology field effect appropriate for this important three-terminal device?
7. Given IDSS 12 mA and VP 6 V, sketch a probable distribution of characteristic curves for
the JFET (similar to Fig. 5.10).
8. In general, comment on the polarity of the various voltages and direction of the currents for an
n-channel JFET versus a p-channel JFET.
§ 5.3 Transfer Characteristics
9. Given the characteristics of Fig. 5.49:
(a) Sketch the transfer characteristics directly from the drain characteristics.
(b) Using Fig. 5.49 to establish the values of IDSS and VP, sketch the transfer characteristics
using Shockley’s equation.
(c) Compare the characteristics of parts (a) and (b). Are there any major differences?
Problems
249
Figure 5.49 Problems 9, 17
10. (a) Given IDSS 12 mA and VP 4 V, sketch the transfer characteristics for the JFET transistor.
(b) Sketch the drain characteristics for the device of part (a).
11. Given IDSS 9 mA and VP 3.5 V, determine ID when:
(a) VGS 0 V.
(b) VGS 2 V.
(c) VGS 3.5 V.
(d) VGS 5 V.
12. Given IDSS 16 mA and VP 5 V, sketch the transfer characteristics using the data points
of Table 5.1. Determine the value of ID at VGS 3 V from the curve, and compare it to the
value determined using Shockley’s equation. Repeat the above for VGS 1 V.
13. A p-channel JFET has device parameters of IDSS 7.5 mA and VP 4 V. Sketch the transfer
characteristics.
14. Given IDSS 6 mA and VP 4.5 V:
(a) Determine ID at VGS 2 and 3.6 V.
(b) Determine VGS at ID 3 and 5.5 mA.
15. Given a Q-point of IDQ 3 mA and VGS 3 V, determine IDSS if VP 6 V.
§ 5.4 Specification Sheets (JFETs)
16. Define the region of operation for the 2N5457 JFET of Fig. .5.18 using the range of IDSS and
VP provided. That is, sketch the transfer curve defined by the maximum IDSS and VP and the
transfer curve for the minimum IDSS and VP. Then, shade in the resulting area between the two
curves.
17. Define the region of operation for the JFET of Fig. 5.49 if VDSmax 25 V and PDmax 120 mW.
§ 5.5 Instrumentation
18. Using the characteristics of Fig. 5.21, determine ID at VGS 0.7 V and VDS 10 V.
19. Referring to Fig. 5.21, is the locus of pinch-off values defined by the region of VDS VP 3 V?
20. Determine VP for the characteristics of Fig. 5.21 using IDSS and ID at some value of VGS. That
is, simply substitute into Shockley’s equation and solve for VP. Compare the result to the assumed value of 3 V from the characteristics.
250
Chapter 5
Field-Effect Transistors
21. Using IDSS 9 mA and VP 3 V for the characteristics of Fig. 5.21, calculate ID at VGS 1 V using Shockley’s equation and compare to the level appearing in Fig. 5.21.
22. (a) Calculate the resistance associated with the JFET of Fig. 5.21 for VGS 0 V from ID 0 to 4 mA.
(b) Repeat part (a) for VGS 0.5 V from ID 0 to 3 mA.
(c) Assigning the label ro to the result of part (a) and rd to that of part (b), use Eq. (5.1) to determine rd and compare to the result of part (b).
§ 5.7 Depletion-Type MOSFET
23. (a) Sketch the basic construction of a p-channel depletion-type MOSFET.
(b) Apply the proper drain-to-source voltage and sketch the flow of electrons for VGS 0 V.
24. In what ways is the construction of a depletion-type MOSFET similar to that of a JFET? In
what ways is it different?
25. Explain in your own words why the application of a positive voltage to the gate of an n-channel depletion-type MOSFET will result in a drain current exceeding IDSS.
26. Given a depletion-type MOSFET with IDSS 6 mA and VP 3 V, determine the drain current at VGS 1, 0, 1, and 2 V. Compare the difference in current levels between 1 and 0 V
with the difference between 1 and 2 V. In the positive VGS region, does the drain current increase at a significantly higher rate than for negative values? Does the ID curve become more
and more vertical with increasing positive values of VGS? Is there a linear or a nonlinear relationship between ID and VGS? Explain.
27. Sketch the transfer and drain characteristics of an n-channel depletion-type MOSFET with
IDSS 12 mA and VP 8 V for a range of VGS VP to VGS 1 V.
28. Given ID 14 mA and VGS 1 V, determine VP if IDSS 9.5 mA for a depletion-type MOSFET.
29. Given ID 4 mA at VGS 2 V, determine IDSS if VP 5 V.
30. Using an average value of 2.9 mA for the IDSS of the 2N3797 MOSFET of Fig. 5.30, determine the level of VGS that will result in a maximum drain current of 20 mA if VP 5 V.
31. If the drain current for the 2N3797 MOSFET of Fig. 5.30 is 8 mA, what is the maximum permissible value of VDS utilizing the maximum power rating?
§ 5.8 Enhancement-Type MOSFET
32. (a) What is the significant difference between the construction of an enhancement-type MOSFET and a depletion-type MOSFET?
(b) Sketch a p-channel enhancement-type MOSFET with the proper biasing applied (VDS 0 V, VGS VT) and indicate the channel, the direction of electron flow, and the resulting
depletion region.
(c) In your own words, briefly describe the basic operation of an enhancement-type MOSFET.
33. (a) Sketch the transfer and drain characteristics of an n-channel enhancement-type MOSFET
if VT 3.5 V and k 0.4 103 A/V2.
(b) Repeat part (a) for the transfer characteristics if VT is maintained at 3.5 V but k is increased
by 100% to 0.8 103 A/V2.
34. (a) Given VGS(Th) 4 V and ID(on) 4 mA at VGS(on) 6 V, determine k and write the general expression for ID in the format of Eq. (5.13).
(b) Sketch the transfer characteristics for the device of part (a).
(c) Determine ID for the device of part (a) at VGS 2, 5, and 10 V.
35. Given the transfer characteristics of Fig. 5.50, determine VT and k and write the general equation for ID.
36. Given k 0.4 103 A/V2 and ID(on) 3 mA with VGS(on) 4 V, determine VT.
37. The maximum drain current for the 2N4351 n-channel enhancement-type MOSFET is 30 mA.
Determine VGS at this current level if k 0.06 103 A/V2 and VT is the maximum value.
Problems
251
Figure 5.50 Problem 35
38. Does the current of an enhancement-type MOSFET increase at about the same rate as a
depletion-type MOSFET for the conduction region? Carefully review the general format of the
equations, and if your mathematics background includes differential calculus, calculate dID /dVGS
and compare its magnitude.
39. Sketch the transfer characteristics of a p-channel enhancement-type MOSFET if VT 5 V
and k 0.45 103 A/V2.
40. Sketch the curve of ID 0.5 103(V 2GS) and ID 0.5 103(VGS 4)2 for VGS from 0 to
10 V. Does VT 4 V have a significant impact on the level of ID for this region?
§ 5.10 VMOS
41. (a) Describe in your own words why the VMOS FET can withstand a higher current and power
rating than the standard construction technique.
(b) Why do VMOS FETs have reduced channel resistance levels?
(c) Why is a positive temperature coefficient desirable?
§ 5.11 CMOS
* 42. (a) Describe in your own words the operation of the network of Fig. 5.44 with Vi 0 V.
(b) If the “on” MOSFET of Fig. 5.44 (with Vi 0 V) has a drain current of 4 mA with
VDS 0.1 V, what is the approximate resistance level of the device? If ID 0.5 A for the
“off” transistor, what is the approximate resistance of the device? Do the resulting resistance levels suggest that the desired output voltage level will result?
43. Research CMOS logic at your local or college library, and describe the range of applications
and basic advantages of the approach.
*Please Note: Asterisks indicate more difficult problems.
252
Chapter 5
Field-Effect Transistors
CHAPTER
FET Biasing
6
6.1 INTRODUCTION
In Chapter 5 we found that the biasing levels for a silicon transistor configuration can
be obtained using the characteristic equations VBE 0.7 V, IC IB, and IC ≅ IE. The
linkage between input and output variables is provided by , which is assumed to be
fixed in magnitude for the analysis to be performed. The fact that beta is a constant
establishes a linear relationship between IC and IB. Doubling the value of IB will double the level of IC, and so on.
For the field-effect transistor, the relationship between input and output quantities
is nonlinear due to the squared term in Shockley’s equation. Linear relationships result in straight lines when plotted on a graph of one variable versus the other, while
nonlinear functions result in curves as obtained for the transfer characteristics of a
JFET. The nonlinear relationship between ID and VGS can complicate the mathematical approach to the dc analysis of FET configurations. A graphical approach may
limit solutions to tenths-place accuracy, but it is a quicker method for most FET amplifiers. Since the graphical approach is in general the most popular, the analysis of
this chapter will have a graphical orientation rather than direct mathematical techniques.
Another distinct difference between the analysis of BJT and FET transistors is
that the input controlling variable for a BJT transistor is a current level, while for the
FET a voltage is the controlling variable. In both cases, however, the controlled variable on the output side is a current level that also defines the important voltage levels of the output circuit.
The general relationships that can be applied to the dc analysis of all FET amplifiers are
IG ≅ 0 A
(6.1)
ID IS
(6.2)
and
For JFETS and depletion-type MOSFETs, Shockley’s equation is applied to relate the input and output quantities:
1 VGS
ID IDSS VP
2
(6.3)
253
For enhancement-type MOSFETs, the following equation is applicable:
ID k(VGS VT)2
(6.4)
It is particularly important to realize that all of the equations above are for the device only! They do not change with each network configuration so long as the device
is in the active region. The network simply defines the level of current and voltage
associated with the operating point through its own set of equations. In reality, the dc
solution of BJT and FET networks is the solution of simultaneous equations established by the device and network. The solution can be determined using a mathematical or graphical approach—a fact to be demonstrated by the first few networks
to be analyzed. However, as noted earlier, the graphical approach is the most popular for FET networks and is employed in this book.
The first few sections of this chapter are limited to JFETs and the graphical approach to analysis. The depletion-type MOSFET will then be examined with its increased range of operating points, followed by the enhancement-type MOSFET.
Finally, problems of a design nature are investigated to fully test the concepts and
procedures introduced in the chapter.
6.2 FIXED-BIAS CONFIGURATION
The simplest of biasing arrangements for the n-channel JFET appears in Fig. 6.1. Referred to as the fixed-bias configuration, it is one of the few FET configurations that
can be solved just as directly using either a mathematical or graphical approach. Both
methods are included in this section to demonstrate the difference between the two
philosophies and also to establish the fact that the same solution can be obtained using either method.
The configuration of Fig. 6.1 includes the ac levels Vi and Vo and the coupling
capacitors (C1 and C2). Recall that the coupling capacitors are “open circuits” for the
dc analysis and low impedances (essentially short circuits) for the ac analysis. The
resistor RG is present to ensure that Vi appears at the input to the FET amplifier for
the ac analysis (Chapter 9). For the dc analysis,
IG ≅ 0 A
VRG IGRG (0 A)RG 0 V
and
The zero-volt drop across RG permits replacing RG by a short-circuit equivalent, as
appearing in the network of Fig. 6.2 specifically redrawn for the dc analysis.
Figure 6.1 Fixed-bias configuration.
254
Chapter 6
FET Biasing
Figure 6.2
Network for dc analysis.
The fact that the negative terminal of the battery is connected directly to the defined positive potential of VGS clearly reveals that the polarity of VGS is directly opposite to that of VGG. Applying Kirchhoff’s voltage law in the clockwise direction of
the indicated loop of Fig. 6.2 will result in
VGG VGS 0
VGS VGG
and
(6.5)
Since VGG is a fixed dc supply, the voltage VGS is fixed in magnitude, resulting in the
notation “fixed-bias configuration.”
The resulting level of drain current ID is now controlled by Shockley’s equation:
V
ID IDSS 1 GS
VP
2
Since VGS is a fixed quantity for this configuration, its magnitude and sign can
simply be substituted into Shockley’s equation and the resulting level of ID calculated.
This is one of the few instances in which a mathematical solution to a FET configuration is quite direct.
A graphical analysis would require a plot of Shockley’s equation as shown in Fig.
6.3. Recall that choosing VGS VP/2 will result in a drain current of IDSS/4 when plotting the equation. For the analysis of this chapter, the three points defined by IDSS,
VP, and the intersection just described will be sufficient for plotting the curve.
ID (mA)
IDSS
IDSS
4
VP
VP
2
0
VGS
Figure 6.3
equation.
Plotting Shockley’s
In Fig. 6.4, the fixed level of VGS has been superimposed as a vertical line at
VGS VGG. At any point on the vertical line, the level of VGS is VGG —the level
of ID must simply be determined on this vertical line. The point where the two curves
ID (mA)
Device
IDSS
Network
Q-point
(solution)
VP
VGSQ = –VGG
ID
0
Q
VGS
Figure 6.4 Finding the solution
for the fixed-bias configuration.
6.2
Fixed-Bias Configuration
255
intersect is the common solution to the configuration—commonly referred to as the
quiescent or operating point. The subscript Q will be applied to drain current and
gate-to-source voltage to identify their levels at the Q-point. Note in Fig. 6.4 that the
quiescent level of ID is determined by drawing a horizontal line from the Q-point to
the vertical ID axis as shown in Fig. 6.4. It is important to realize that once the network of Fig. 6.1 is constructed and operating, the dc levels of ID and VGS that will be
measured by the meters of Fig. 6.5 are the quiescent values defined by Fig. 6.4.
Figure 6.5 Measuring the quiescent values of ID and VGS.
The drain-to-source voltage of the output section can be determined by applying
Kirchhoff’s voltage law as follows:
VDS ID RD VDD 0
VDS VDD IDRD
and
(6.6)
Recall that single-subscript voltages refer to the voltage at a point with respect to
ground. For the configuration of Fig. 6.2,
VS 0 V
(6.7)
Using double-subscript notation:
VDS VD VS
VD VDS VS VDS 0 V
or
VD VDS
and
In addition,
(6.8)
VGS VG VS
VG VGS VS VGS 0 V
or
VG VGS
and
(6.9)
The fact that VD VDS and VG VGS is fairly obvious from the fact that VS 0 V, but the derivations above were included to emphasize the relationship that exists
between double-subscript and single-subscript notation. Since the configuration requires two dc supplies, its use is limited and will not be included in the forthcoming
list of the most common FET configurations.
256
Chapter 6
FET Biasing
EXAMPLE 6.1
Determine the following for the network of Fig. 6.6.
(a) VGSQ.
16 V
(b) IDQ.
(c) VDS.
(d) VD.
2 kΩ
(e) VG.
(f) VS.
D
G
I DSS = 10 mA
VP = –8 V
+
1 MΩ
–
VGS
–
S
2V
+
Figure 6.6 Example 6.1.
Solution
Mathematical Approach:
(a) VGSQ VGG 2 V
V 2
2 V 2
(b) IDQ IDSS 1 GS 10 mA 1 VP
8 V
2
10 mA(1 0.25) 10 mA(0.75)2 10 mA(0.5625)
5.625 mA
(c) VDS VDD IDRD 16 V (5.625 mA)(2 k)
16 V 11.25 V 4.75 V
(d) VD VDS 4.75 V
(e) VG VGS 2 V
(f) VS 0 V
Graphical Approach:
The resulting Shockley curve and the vertical line at VGS 2 V are provided in Fig.
6.7. It is certainly difficult to read beyond the second place without significantly inID (mA)
IDSS = 10 mA
9
8
7
Q-point
6
5
I D = 5.6 mA
Q
4
3
2
IDSS = 2.5 mA
4
1
–8 –7 – 6 – 5 – 4 – 3 –2 –1
0
VGS
VP = –8 V VP
V
=
–V
=
–2
V
GSQ
GG
= –4 V
2
Figure 6.7 Graphical solution
for the network of Fig. 6.6.
6.2
Fixed-Bias Configuration
257
creasing the size of the figure, but a solution of 5.6 mA from the graph of Fig. 6.7 is
quite acceptable. Therefore, for part (a),
VGSQ VGG 2 V
(b) IDQ 5.6 mA
(c) VDS VDD IDRD 16 V (5.6 mA)(2 k)
16 V 11.2 V 4.8 V
(d) VD VDS 4.8 V
(e) VG VGS 2 V
(f) VS 0 V
The results clearly confirm the fact that the mathematical and graphical approaches
generate solutions that are quite close.
6.3 SELF-BIAS CONFIGURATION
The self-bias configuration eliminates the need for two dc supplies. The controlling
gate-to-source voltage is now determined by the voltage across a resistor RS introduced in the source leg of the configuration as shown in Fig. 6.8.
Figure 6.8
figuration.
JFET self-bias con-
For the dc analysis, the capacitors can again be replaced by “open circuits” and
the resistor RG replaced by a short-circuit equivalent since IG 0 A. The result is the
network of Fig. 6.9 for the important dc analysis.
The current through RS is the source current IS, but IS ID and
VRS IDRS
For the indicated closed loop of Fig. 6.9, we find that
VGS VRS 0
and
VGS VRS
or
VGS IDRS
(6.10)
Figure 6.9 DC analysis of the
self-bias configuration.
Note in this case that VGS is a function of the output current ID and not fixed in magnitude as occurred for the fixed-bias configuration.
258
Chapter 6
FET Biasing
Equation (6.10) is defined by the network configuration, and Shockley’s equation
relates the input and output quantities of the device. Both equations relate the same
two variables, permitting either a mathematical or graphical solution.
A mathematical solution could be obtained simply by substituting Eq. (6.10) into
Shockley’s equation as shown below:
V
ID IDSS 1 GS
VP
2
I RS
IDSS 1 D
VP
or
2
IDRS
ID IDSS 1 VP
2
By performing the squaring process indicated and rearranging terms, an equation of
the following form can be obtained:
ID2 K1ID K2 0
The quadratic equation can then be solved for the appropriate solution for ID.
The sequence above defines the mathematical approach. The graphical approach
requires that we first establish the device transfer characteristics as shown in Fig. 6.10.
Since Eq. (6.10) defines a straight line on the same graph, let us now identify two
points on the graph that are on the line and simply draw a straight line between the
two points. The most obvious condition to apply is ID 0 A since it results in
VGS IDRS (0 A)RS 0 V. For Eq. (6.10), therefore, one point on the straight
line is defined by ID 0 A and VGS 0 V, as appearing on Fig. 6.10.
Figure 6.10 Defining a point
on the self-bias line.
The second point for Eq. (6.10) requires that a level of VGS or ID be chosen and
the corresponding level of the other quantity be determined using Eq. (6.10). The resulting levels of ID and VGS will then define another point on the straight line and permit an actual drawing of the straight line. Suppose, for example, that we choose a
level of ID equal to one-half the saturation level. That is,
I SS
ID D
2
then
IDSSRS
VGS IDRS 2
The result is a second point for the straight-line plot as shown in Fig. 6.11. The straight
line as defined by Eq. (6.10) is then drawn and the quiescent point obtained at the in6.3
Self-Bias Configuration
259
ID
IDSS
IDSS
2
Q-point
ID
Q
VP
VGSQ
I
R
VGS = _ DSS S
2
0
VGS
Figure 6.11 Sketching the selfbias line.
tersection of the straight-line plot and the device characteristic curve. The quiescent
values of ID and VGS can then be determined and used to find the other quantities of
interest.
The level of VDS can be determined by applying Kirchhoff’s voltage law to the
output circuit, with the result that
VRS VDS VRD VDD 0
VDS VDD VRS VRD VDD ISRS IDRD
and
ID IS
but
and
VDS VDD ID(RS RD)
(6.11)
VS IDRS
(6.12)
VG 0 V
(6.13)
VD VDS VS VDD VRD
(6.14)
In addition:
and
EXAMPLE 6.2
Determine the following for the network of Fig. 6.12.
(a) VGSQ.
(b) IDQ.
(c) VDS.
(d) VS.
(e) VG.
(f ) VD.
Figure 6.12 Example 6.2.
260
Chapter 6
FET Biasing
Solution
(a) The gate-to-source voltage is determined by
VGS IDRS
Choosing ID 4 mA, we obtain
VGS (4 mA)(1 k) 4 V
The result is the plot of Fig. 6.13 as defined by the network.
ID = 8 mA, VGS = –8 V
ID (mA)
8
7
ID = 4 mA, VGS = – 4V
Network
6
5
4
3
2
1 V = 0 V, I = 0 mA
GS
D
– 8 –7 – 6 – 5 – 4 – 3 –2 – 1
0
VGS (V)
Figure 6.13 Sketching the selfbias line for the network of Fig.
6.12.
If we happen to choose ID 8 mA, the resulting value of VGS would be 8 V, as
shown on the same graph. In either case, the same straight line will result, clearly
demonstrating that any appropriate value of ID can be chosen as long as the corresponding value of VGS as determined by Eq. (6.10) is employed. In addition, keep in
mind that the value of VGS could be chosen and the value of ID calculated with the
same resulting plot.
For Shockley’s equation, if we choose VGS VP/2 3 V, we find that ID IDSS/4 8 mA/4 2 mA, and the plot of Fig. 6.14 will result, representing the characteristics of the device. The solution is obtained by superimposing the network characteristics defined by Fig. 6.13 on the device characteristics of Fig. 6.14 and finding
the point of intersection of the two as indicated on Fig. 6.15. The resulting operating
point results in a quiescent value of gate-to-source voltage of
VGSQ 2.6 V
ID (mA)
8
7
6
5
4
3
Q-point
2
I D = 2.6 mA
Q
1
– 6 – 5 – 4 – 3 –2 –1
0
VGS (V)
VGSQ = – 2.6 V
Figure 6.14 Sketching the device characteristics for the JFET of Fig. 6.12.
Figure 6.15 Determining the Q-point for the
network of Fig. 6.12.
6.3
Self-Bias Configuration
261
(b) At the quiescent point:
IDQ 2.6 mA
(c) Eq. (6.11): VDS VDD ID(RS RD)
20 V (2.6 mA)(1 k 3.3 k)
20 V 11.18 V
8.82 V
(d) Eq. (6.12): VS IDRS
(2.6 mA)(1 k)
2.6 V
(e) Eq. (6.13): VG 0 V
(f) Eq. (6.14): VD VDS VS 8.82 V 2.6 V 11.42 V
or
VD VDD IDRD 20 V (2.6 mA)(3.3 k) 11.42 V
EXAMPLE 6.3
Find the quiescent point for the network of Fig. 6.12 if:
(a) RS 100 .
(b) RS 10 k.
Solution
Note Fig. 6.16.
ID (mA)
8
RS = 100 Ω
I D = 4 mA, VGS = – 0.4 V
7
Q-point
6
I D ≅ 6.4 mA
Q
5
4
RS = 10 kΩ
VGS = –4 V, ID = 0.4 mA
3
2
Q-point
1
– 6 – 5 – 4 – 3 –2 –1
0
VGS (V)
VGSQ ≅ – 4.6 V
Figure 6.16 Example 6.3.
(a) With the ID scale,
IDQ ≅ 6.4 mA
From Eq. (6.10),
VGSQ ≅ 0.64 V
(b) With the VGS scale,
VGSQ ≅ 4.6 V
From Eq. (6.10),
IDQ ≅ 0.46 mA
In particular, note how lower levels of RS bring the load line of the network closer
to the ID axis while increasing levels of RS bring the load line closer to the VGS axis.
262
Chapter 6
FET Biasing
Determine the following for the common-gate configuration of Fig. 6.17.
(a) VGSQ.
(b) IDQ.
(c) VD.
(d) VG.
(e) VS.
(f) VDS.
EXAMPLE 6.4
Figure 6.17 Example 6.4.
Solution
The grounded gate terminal and the location of the input establish strong similarities
with the common-base BJT amplifier. Although different in appearance from the basic structure of Fig. 6.8, the resulting dc network of Fig. 6.18 has the same basic structure as Fig. 6.9. The dc analysis can therefore proceed in the same manner as recent
examples.
(a) The transfer characteristics and load line appear in Fig. 6.19. In this case, the second point for the sketch of the load line was determined by choosing (arbitrarily)
ID 6 mA and solving for VGS. That is,
VGS IDRS (6 mA)(680 ) 4.08 V
as shown in Fig. 6.19. The device transfer curve was sketched using
I SS
12 mA
3 mA
ID D
4
4
ID (mA)
Figure 6.18 Sketching the dc
equivalent of the network of Fig.
6.17.
12 I DSS
11
10
9
8
7
6
5
4
Q-point
3
I D ≅ 3.8 mA
Q
2
1
–6
VP
–5
–4
–3
–2
–1
VGSQ ≅ –2.6 V
0
Figure 6.19 Determining the
Q-point for the network of Fig.
6.17.
6.3
Self-Bias Configuration
263
and the associated value of VGS:
VP
6V
VGS 3 V
2
2
as shown on Fig. 6.19. Using the resulting quiescent point of Fig. 6.19 results in
VGSQ ≅ 2.6 V
(b) From Fig. 6.19,
IDQ ≅ 3.8 mA
(c) VD VDD IDRD
12 V (3.8 mA)(1.5 k) 12 V 5.7 V
6.3 V
(d) VG 0 V
(e) VS IDRS (3.8 mA)(680 )
2.58 V
(f) VDS VD VS
6.3 V 2.58 V
3.72 V
6.4 VOLTAGE-DIVIDER BIASING
The voltage-divider bias arrangement applied to BJT transistor amplifiers is also applied to FET amplifiers as demonstrated by Fig. 6.20. The basic construction is exactly the same, but the dc analysis of each is quite different. IG 0 A for FET amplifiers, but the magnitude of IB for common-emitter BJT amplifiers can affect the dc
levels of current and voltage in both the input and output circuits. Recall that IB provided the link between input and output circuits for the BJT voltage-divider configuration while VGS will do the same for the FET configuration.
The network of Fig. 6.20 is redrawn as shown in Fig. 6.21 for the dc analysis.
Note that all the capacitors, including the bypass capacitor CS, have been replaced by
an “open-circuit” equivalent. In addition, the source VDD was separated into two equivVDD
R1
VDD
VDD
RD
R1
ID
IG ≅ 0 A
VG
+
+
R2
–
+
R2
VG
–
VGS
–
+
VRS
IS
RS
–
Figure 6.21 Redrawn network of Fig. 6.20 for dc analysis.
Figure 6.20 Voltage-divider bias arrangement.
264
Chapter 6
FET Biasing
alent sources to permit a further separation of the input and output regions of the network. Since IG 0 A, Kirchhoff’s current law requires that IR1 IR2 and the series
equivalent circuit appearing to the left of the figure can be used to find the level of
VG. The voltage VG, equal to the voltage across R2, can be found using the voltagedivider rule as follows:
R2VDD
VG R1 R2
(6.15)
Applying Kirchhoff’s voltage law in the clockwise direction to the indicated loop
of Fig. 6.21 will result in
VG VGS VRS 0
and
VGS VG VRS
Substituting VRS ISRS ID RS, we have
VGS VG IDRS
(6.16)
The result is an equation that continues to include the same two variables appearing in Shockley’s equation: VGS and ID. The quantities VG and RS are fixed by
the network construction. Equation (6.16) is still the equation for a straight line, but
the origin is no longer a point in the plotting of the line. The procedure for plotting
Eq. (6.16) is not a difficult one and will proceed as follows. Since any straight line
requires two points to be defined, let us first use the fact that anywhere on the horizontal axis of Fig. 6.22 the current ID 0 mA. If we therefore select ID to be 0 mA,
we are in essence stating that we are somewhere on the horizontal axis. The exact location can be determined simply by substituting ID 0 mA into Eq. (6.16) and finding the resulting value of VGS as follows:
VGS VG IDRS
VG (0 mA)RS
and
VGS VGID0 mA
(6.17)
The result specifies that whenever we plot Eq. (6.16), if we choose ID 0 mA, the
value of VGS for the plot will be VG volts. The point just determined appears in Fig.
6.22.
Figure 6.22 Sketching the network equation for the voltage-divider configuration.
6.4
Voltage-Divider Biasing
265
For the other point, let us now employ the fact that at any point on the vertical
axis VGS 0 V and solve for the resulting value of ID:
VGS VG IDRS
0 V VG IDRS
VG
ID RS
and

VGS 0 V
(6.18)
The result specifies that whenever we plot Eq. (6.16), if VGS 0 V, the level of ID is
determined by Eq. (6.18). This intersection also appears on Fig. 6.22.
The two points defined above permit the drawing of a straight line to represent
Eq. (6.16). The intersection of the straight line with the transfer curve in the region
to the left of the vertical axis will define the operating point and the corresponding
levels of ID and VGS.
Since the intersection on the vertical axis is determined by ID VG/RS and VG is
fixed by the input network, increasing values of RS will reduce the level of the ID intersection as shown in Fig. 6.23. It is fairly obvious from Fig. 6.23 that:
Increasing values of RS result in lower quiescent values of ID and more negative values of VGS.
Figure 6.23 Effect of RS on the resulting Q-point.
Once the quiescent values of IDQ and VGSQ are determined, the remaining network
analysis can be performed in the usual manner. That is,
266
Chapter 6
FET Biasing
VDS VDD ID(RD RS)
(6.19)
VD VDD IDRD
(6.20)
VS IDRS
(6.21)
V
IR1 IR2 DD
R1 R2
(6.22)
Determine the following for the network of Fig. 6.24.
(a) IDQ and VGSQ.
(b) VD.
(c) VS.
(d) VDS.
(e) VDG.
EXAMPLE 6.5
Figure 6.24 Example 6.5.
Solution
(a) For the transfer characteristics, if ID IDSS/4 8 mA/4 2 mA, then VGS VP/2 4 V/2 2 V. The resulting curve representing Shockley’s equation appears in Fig. 6.25. The network equation is defined by
R2VDD
VG R1 R2
(270 k)(16 V)
2.1 M 0.27 M
1.82 V
VGS VG IDRS
and
1.82 V ID(1.5 k)
When ID 0 mA:
VGS 1.82 V
ID (mA)
8 (IDSS )
7
6
5
4
3
Q-point
2
1
–4
(VP)
–3
–2
–1
0
VGSQ = –1.8 V
I D = 2.4 mA
Q
I D =1.21 mA ( VGS = 0 V)
1
2
3
VG = 1.82 V
( I D = 0 mA )
Figure 6.25 Determining the
Q-point for the network of Fig.
6.24.
6.4
Voltage-Divider Biasing
267
When VGS 0 V:
1.82 V
ID 1.21 mA
1.5 k
The resulting bias line appears on Fig. 6.25 with quiescent values of
IDQ 2.4 mA
VGSQ 1.8 V
and
(b) VD VDD IDRD
16 V (2.4 mA)(2.4 k)
10.24 V
(c) VS IDRS (2.4 mA)(1.5 k)
3.6 V
(d) VDS VDD ID(RD RS)
16 V (2.4 mA)(2.4 k 1.5 k)
6.64 V
or VDS VD VS 10.24 V 3.6 V
6.64 V
(e) Although seldom requested, the voltage VDG can easily be determined using
VDG VD VG
10.24 V 1.82 V
8.42 V
Although the basic construction of the network in the next example is quite different from the voltage-divider bias arrangement, the resulting equations require a solution very similar to that just described. Note that the network employs a supply at
the drain and source.
EXAMPLE 6.6
Determine the following for the network of Fig. 6.26.
(a) IDQ and VGSQ.
(b) VDS.
VDD = 20 V
(c) VD.
ID
(d) VS.
R D = 1.8 k Ω
I DSS = 9 mA
VP = –3 V
RS = 1.5 kΩ
VSS = –10 V
268
Chapter 6
FET Biasing
Figure 6.26 Example 6.6.
Solution
(a) An equation for VGS in terms of ID is obtained by applying Kirchhoff’s voltage
law to the input section of the network as redrawn in Fig. 6.27.
VGS ISRS VSS 0
or
VGS VSS ISRS
but
IS ID
and
VGS VSS IDRS
(6.23)
The result is an equation very similar in format to Eq. (6.16) that can be superimposed on the transfer characteristics using the procedure described for Eq. (6.16).
That is, for this example,
Figure 6.27 Determining the
network equation for the configuration of Fig. 6.26.
VGS 10 V ID(1.5 k)
For ID 0 mA,
VGS VSS 10 V
For VGS 0 V,
0 10 V ID(1.5 k)
and
10 V
ID 6.67 mA
1.5 k
The resulting plot points are identified on Fig. 6.28.
Figure 6.28 Determining the
Q-point for the network of Fig.
6.26.
The transfer characteristics are sketched using the plot point established by VGS VP/2 3 V/2 1.5 V and ID IDSS/4 9 mA/4 2.25 mA, as also appearing
on Fig. 6.28. The resulting operating point establishes the following quiescent levels:
IDQ 6.9 mA
VGSQ 0.35 V
(b) Applying Kirchhoff’s voltage law to the output side of Fig. 6.26 will result in
VSS ISRS VDS IDRD VDD 0
6.4
Voltage-Divider Biasing
269
Substituting IS ID and rearranging gives
VDS VDD VSS ID(RD RS)
(6.24)
which for this example results in
VDS 20 V 10 V (6.9 mA)(1.8 k 1.5 k)
30 V 22.77 V
7.23 V
(c) VD VDD IDRD
20 V (6.9 mA)(1.8 k) 20 V 12.42 V
7.58 V
(d) VDS VD VS
or VS VD VDS
7.58 V 7.23 V
0.35 V
6.5 DEPLETION-TYPE MOSFETs
The similarities in appearance between the transfer curves of JFETs and depletiontype MOSFETs permit a similar analysis of each in the dc domain. The primary difference between the two is the fact that depletion-type MOSFETs permit operating
points with positive values of VGS and levels of ID that exceed IDSS. In fact, for all
the configurations discussed thus far, the analysis is the same if the JFET is replaced
by a depletion-type MOSFET.
The only undefined part of the analysis is how to plot Shockley’s equation for
positive values of VGS. How far into the region of positive values of VGS and values
of ID greater than IDSS does the transfer curve have to extend? For most situations,
this required range will be fairly well defined by the MOSFET parameters and the
resulting bias line of the network. A few examples will reveal the impact of the change
in device on the resulting analysis.
EXAMPLE 6.7
For the n-channel depletion-type MOSFET of Fig. 6.29, determine:
(a) IDQ and VGSQ.
(b) VDS.
Figure 6.29 Example 6.7.
270
Solution
(a) For the transfer characteristics, a plot point is defined by ID IDSS/4 6 mA/4 1.5 mA and VGS VP/2 3 V/2 1.5 V. Considering the level of VP and
the fact that Shockley’s equation defines a curve that rises more rapidly as VGS
becomes more positive, a plot point will be defined at VGS 1 V. Substituting
into Shockley’s equation yields
V
ID IDSS 1 GS
VP
2
1 V
6 mA 1 3 V
2
1
6 mA 1 3
2
6 mA(1.778)
10.67 mA
The resulting transfer curve appears in Fig. 6.30. Proceeding as described for JFETs,
we have:
Eq. (6.15):
10 M(18 V)
VG 1.5 V
10 M 110 M
Eq. (6.16):
VGS VG IDRS 1.5 V ID(750 )
Figure 6.30 Determining the
Q-point for the network of Fig.
6.29.
Setting ID 0 mA results in
VGS VG 1.5 V
Setting VGS 0 V yields
VG
1.5 V
ID 2 mA
RS
750 The plot points and resulting bias line appear in Fig. 6.30. The resulting operating
point:
IDQ 3.1 mA
VGSQ 0.8 V
6.5
Depletion-Type MOSFETs
271
(b) Eq. (6.19): VDS VDD ID(RD RS)
18 V (3.1 mA)(1.8 k 750 )
≅ 10.1 V
EXAMPLE 6.8
Repeat Example 6.7 with RS 150 .
Solution
(a) The plot points are the same for the transfer curve as shown in Fig. 6.31. For the
bias line,
VGS VG IDRS 1.5 V ID(150 )
Figure 6.31 Example 6.8.
Setting ID 0 mA results in
VGS 1.5 V
Setting VGS 0 V yields
VG
1.5 V
ID 10 mA
RS
150 The bias line is included on Fig. 6.31. Note in this case that the quiescent point results in a drain current that exceeds IDSS, with a positive value for VGS. The result:
IDQ 7.6 mA
VGSQ 0.35 V
(b) Eq. (6.19): VDS VDD ID(RD RS)
18 V (7.6 mA)(1.8 k 150 )
3.18 V
272
Chapter 6
FET Biasing
EXAMPLE 6.9
Determine the following for the network of Fig. 6.32.
(a) IDQ and VGSQ.
(b) VD.
20 V
6.2 kΩ
Vo
I DSS = 8 mA
VP = – 8 V
Vi
1 MΩ
2.4 kΩ
Figure 6.32 Example 6.9.
Solution
(a) The self-bias configuration results in
VGS IDRS
as obtained for the JFET configuration, establishing the fact that VGS must be less
than zero volts. There is therefore no requirement to plot the transfer curve for positive values of VGS, although it was done on this occasion to complete the transfer
characteristics. A plot point for the transfer characteristics for VGS 0 V is
I SS
8 mA
ID D
2 mA
4
4
VP
8 V
4 V
VGS 2
2
and
and for VGS
0 V, since VP 8 V, we will choose
VGS 2 V
and
V
ID IDSS 1 GS
VP
2
2 V
8 mA 1 8 V
2
12.5 mA
The resulting transfer curve appears in Fig. 6.33. For the network bias line, at VGS 0 V, ID 0 mA. Choosing VGS 6 V gives
V
6 V
ID GS 2.5 mA
RS
2.4 k
The resulting Q-point:
IDQ 1.7 mA
VGSQ 4.3 V
(b) VD VDD IDRD
20 V (1.7 mA)(6.2 k)
9.46 V
6.5
Depletion-Type MOSFETs
273
Figure 6.33 Determining the Qpoint for the network of Fig.
6.32.
The example to follow employs a design that can also be applied to JFET transistors. At first impression it appears rather simplistic, but in fact it often causes some
confusion when first analyzed due to the special point of operation.
EXAMPLE 6.10
Determine VDS for the network of Fig. 6.34.
Solution
The direct connection between the gate and source terminals requires that
VGS 0 V
Since VGS is fixed at 0 V, the drain current must be IDSS (by definition). In other
words,
VGSQ 0 V
IDQ 10 mA
and
There is therefore no need to draw the transfer curve and
VD VDD IDRD 20 V (10 mA)(1.5 k)
20 V 15 V
5V
Figure 6.34 Example 6.10.
6.6 ENHANCEMENT-TYPE MOSFETs
The transfer characteristics of the enhancement-type MOSFET are quite different from
those encountered for the JFET and depletion-type MOSFETs, resulting in a graphical solution quite different from the preceding sections. First and foremost, recall that
for the n-channel enhancement-type MOSFET, the drain current is zero for levels of
gate-to-source voltage less than the threshold level VGS(Th), as shown in Fig. 6.35. For
levels of VGS greater than VGS(Th), the drain current is defined by
274
Chapter 6
FET Biasing
ID (mA)
ID2
ID = k (VGS – VGS(Th) )2
ID (on)
ID1
VGS(Th)
VGS1
ID = 0 mA
VGS(on)
VGS2
VGS
Figure 6.35 Transfer characteristics of an n-channel enhancementtype MOSFET.
ID k(VGS VGS(Th))2
(6.25)
Since specification sheets typically provide the threshold voltage and a level of drain
current (ID(on)) and its corresponding level of VGS(on), two points are defined immediately as shown in Fig. 6.35. To complete the curve, the constant k of Eq. (6.25) must
be determined from the specification sheet data by substituting into Eq. (6.25) and
solving for k as follows:
ID k(VGS VGS(Th))2
ID(on) k(VGS(on) VGS(Th))2
and
ID(on)
k (VGS(on) VGS(Th))2
(6.26)
Once k is defined, other levels of ID can be determined for chosen values of VGS. Typically, a point between VGS(Th) and VGS(on) and one just greater than VGS(on) will provide a sufficient number of points to plot Eq. (6.25) (note ID1 and ID2 on Fig.
6.35).
Feedback Biasing Arrangement
A popular biasing arrangement for enhancement-type MOSFETs is provided in Fig.
6.36. The resistor RG brings a suitably large voltage to the gate to drive the MOSFET
“on.” Since IG 0 mA and VRG 0 V, the dc equivalent network appears as shown
in Fig. 6.37.
A direct connection now exists between drain and gate, resulting in
VD VG
and
VDS VGS
(6.27)
6.6
Enhancement-Type MOSFETs
275
Figure 6.37 DC equivalent of
the network of Fig. 6.36.
Figure 6.36 Feedback biasing arrangement.
For the output circuit,
VDS VDD IDRD
which becomes the following after substituting Eq. (6.27):
VGS VDD IDRD
(6.28)
The result is an equation that relates the same two variables as Eq. (6.25), permitting
the plot of each on the same set of axes.
Since Eq. (6.28) is that of a straight line, the same procedure described earlier can
be employed to determine the two points that will define the plot on the graph. Substituting ID 0 mA into Eq. (6.28) gives
VGS VDDID 0 mA
(6.29)
Substituting VGS 0 V into Eq. (6.28), we have
V D
ID D
RD

VGS 0 V
(6.30)
The plots defined by Eqs. (6.25) and (6.28) appear in Fig. 6.38 with the resulting operating point.
Figure 6.38 Determining the Qpoint for the network of Fig.
6.36.
276
Chapter 6
FET Biasing
Determine IDQ and VDSQ for the enhancement-type MOSFET of Fig. 6.39.
EXAMPLE 6.11
Figure 6.39 Example 6.11.
Solution
Plotting the Transfer Curve:
Two points are defined immediately as shown in Fig. 6.40. Solving for k:
Eq. (6.26):
ID(on)
k (VGS(on) VGS(Th))2
6 mA
6 103
2 A/v2
(8 V 3 V)
25
0.24 103 A/V2
For VGS 6 V (between 3 and 8 V):
ID 0.24
103(6 V 3 V)2 0.24
103(9)
2.16 mA
Figure 6.40 Plotting the transfer curve for the MOSFET of Fig.
6.39.
6.6
Enhancement-Type MOSFETs
277
as shown on Fig. 6.40. For VGS 10 V (slightly greater than VGS(Th)):
103(10 V 3 V)2 0.24
ID 0.24
103(49)
11.76 mA
as also appearing on Fig. 6.40. The four points are sufficient to plot the full curve for
the range of interest as shown in Fig. 6.40.
For the Network Bias Line:
VGS VDD IDRD
12 V ID(2 k)
Eq. (6.29):
VGS VDD 12 VID 0 mA
Eq. (6.30):
V D
12 V
ID D
6 mAVGS 0 V
RD
2 k
The resulting bias line appears in Fig. 6.41.
At the operating point:
IDQ 2.75 mA
and
VGSQ 6.4 V
with
VDSQ VGSQ 6.4 V
ID = mA
12
11
10
9
8
7
VDD
6
RD
5
4
I D = 2.75 mA
Q
3
Q-point
2
1
0
1
2
3
4
5
6
7
9 10 11 12
(VDD)
VGSQ = 6.4 V
8
VGS
Figure 6.41 Determining the Q-point for the network of Fig. 6.39.
Voltage-Divider Biasing Arrangement
A second popular biasing arrangement for the enhancement-type MOSFET appears
in Fig. 6.42. The fact that IG 0 mA results in the following equation for VGG as derived from an application of the voltage-divider rule:
R2VDD
VG R1 R2
Figure 6.42 Voltage-divider
biasing arrangement for an nchannel enhancement MOSFET.
278
Chapter 6
FET Biasing
(6.31)
Applying Kirchhoff’s voltage law around the indicated loop of Fig. 6.42 will result in
VG VGS VRS 0
and
VGS VG VRS
or
VGS VG IDRS
(6.32)
For the output section:
VRS VDS VRD VDD 0
VDS VDD VRS VRD
and
VDS VDD ID(RS RD)
or
(6.33)
Since the characteristics are a plot of ID versus VGS and Eq. (6.32) relates the same
two variables, the two curves can be plotted on the same graph and a solution determined at their intersection. Once IDQ and VGSQ are known, all the remaining quantities of the network such as VDS, VD, and VS can be determined.
Determine IDQ, VGSQ, and VDS for the network of Fig. 6.43.
EXAMPLE 6.12
Figure 6.43 Example 6.12.
Solution
Network:
Eq. (6.31):
R2VDD
(18 M)(40 V)
VG 18 V
R1 R2
22 M 18 M
Eq. (6.32):
VGS VG IDRS 18 V ID(0.82 k)
When ID 0 mA,
VGS 18 V (0 mA)(0.82 k) 18 V
as appearing on Fig. 6.44. When VGS 0 V,
VGS 18 V ID(0.82 k)
0 18 V ID(0.82 k)
18 V
ID 21.95 mA
0.82 k
as appearing on Fig. 6.44.
6.6
Enhancement-Type MOSFETs
279
ID (mA)
30
VG
= 21.95
RS
20
10
I D ≅ 6.7 mA
Q-point
Q
0
5
VGS (Th)
10
15
20
VGSQ = 12.5 V VG = 18 V
25
VGS
Figure 6.44 Determining the Q-point for the network of Example 6.12.
Device:
VGS(Th) 5 V,
Eq. (6.26):
ID(on) 3 mA with VGS(on) 10 V
ID(on)
k (VGS(on) VGS(Th))2
3 mA
0.12
(10 V 5 V)2
103 A/V2
ID k(VGS VGS(Th))2
and
0.12
103(VGS 5)2
which is plotted on the same graph (Fig. 6.44). From Fig. 6.44,
IDQ ≅ 6.7 mA
VGSQ 12.5 V
Eq. (6.33):
VDS VDD ID(RS RD)
40 V (6.7 mA)(0.82 k 3.0 k)
40 V 25.6 V
14.4 V
6.7 SUMMARY TABLE
Now that the most popular biasing arrangements for the various FETs have been introduced, Table 6.1 reviews the basic results and demonstrates the similarity in approach for a number of configurations. It also reveals that the general analysis of dc
configurations for FETs is not overly complex. Once the transfer characteristics are
established, the network self-bias line can be drawn and the Q-point determined at
the intersection of the device transfer characteristic and the network bias curve. The
remaining analysis is simply an application of the basic laws of circuit analysis.
280
Chapter 6
FET Biasing
TABLE 6.1 FET Bias Configurations
Type
Configuration
VDD
RD
JFET
Fixed-bias
RG
VGG –
Pertinent Equations
Graphical Solution
ID
IDSS
VGSQ VGG
VDS VDD IDRS
Q-point
+
VP VGG 0
VDD
RD
JFET
Self-bias
RG
ID
IDSS
VGS IDRS
VDS VDD ID(RD RS)
R1
R2
I'D
Q-point
RS
VDD
JFET
Voltage-divider
bias
VP V'
0
GS
RD
R2VDD
VG R1 R2
RS
VGS VG IDRS
VDS VDD ID(RD RS)
RS
–VSS
VDD
RD
JFET
(VGSQ 0 V)
VG
RS
Q-point
VP
0
VDD
RG
RS
VG VGS
ID
IDSS
VGS VSS IDRS
VDS VDD VSS ID(RD RS)
VSS
RS
Q-point
VP
0
Q-point
ID
IDSS
VGSQ 0 V
IDQ IDSS
VGS IDRS
VD VDD
VS IDRS
VDS VDD ISRS
VSS VGS
VGS = 0 V
Q
VP
JFET
(RD 0 )
VGS
ID
IDSS
VDD
RD
JFET
Common-gate
VGS
0
VGS
ID
IDSS
I'D
Q-point
VP V'GS 0
ID
Q-point
VDD
RD
Depletion-type
MOSFET
Fixed-bias
VGSQ VGG
VDS VDD IDRS
RG
IDSS
VGG
Depletion-type
MOSFET
Voltage-divider
bias
Enhancement
type MOSFETFeedback
configuration
VP
R1
RD
R2
RS
VDD
RG
R1
R2
RD
R2VDD
VG R1 R2
RS
VGS VG IDRS
VGS
ID
Q-point
0
VG VGS
ID
ID(on)
Q-point
0
VDD
Enhancementtype MOSFET
Voltage-divider
bias
VP
VDD
RD
VGS VDS
VGS VDD IDRD
0 VGG
VG
RS
IDSS
R2VDD
VG R1 R2
VGS VG ISRS
VDS VDD ID(RD RS)
VDD
RD
VGS
VG
RS
VGS(Th)
VGS(on)
VDD VGS
ID
Q-point
0
VGS(Th)
VG VGS
281
6.8 COMBINATION NETWORKS
Now that the dc analysis of a variety of BJT and FET configurations is established,
the opportunity to analyze networks with both types of devices presents itself. Fundamentally, the analysis simply requires that we first approach the device that will
provide a terminal voltage or current level. The door is then usually open to calculate other quantities and concentrate on the remaining unknowns. These are usually
particularly interesting problems due to the challenge of finding the opening and then
using the results of the past few sections and Chapter 5 to find the important quantities for each device. The equations and relationships used are simply those we have
now employed on more than one occasion—no need to develop any new methods of
analysis.
EXAMPLE 6.13
Determine the levels of VD and VC for the network of Fig. 6.45.
Figure 6.45 Example 6.13.
Solution
From past experience we now realize that VGS is typically an important quantity to
determine or write an equation for when analyzing JFET networks. Since VGS is a
level for which an immediate solution is not obvious, let us turn our attention to the
transistor configuration. The voltage-divider configuration is one where the approximate technique can be applied (RE (180 1.6 k) 288 k 10R2 240 k),
permitting a determination of VB using the voltage-divider rule on the input circuit.
For VB:
24 k(16 V)
VB 3.62 V
82 k 24 k
Using the fact that VBE 0.7 V results in
VE VB VBE 3.62 V 0.7 V
2.92 V
282
Chapter 6
FET Biasing
and
V E
VE
2.92 V
IE R
1.825 mA
RE
RE
1.6 k
with
IC ≅ IE 1.825 mA
Continuing, we find for this configuration that
ID IS IC
VD 16 V ID(2.7 k)
and
16 V (1.825 mA)(2.7 k) 16 V 4.93 V
11.07 V
The question of how to determine VC is not as obvious. Both VCE and VDS are unknown quantities preventing us from establishing a link between VD and VC or from
VE to VD. A more careful examination of Fig. 6.45 reveals that VC is linked to VB by
VGS (assuming that VRG 0 V). Since we know VB if we can find VGS, VC can be
determined from
VC VB VGS
The question then arises as to how to find the level of VGSQ from the quiescent
value of ID. The two are related by Shockley’s equation:
V S
IDQ IDSS 1 G
VP
Q
2
and VGSQ could be found mathematically by solving for VGSQ and substituting numerical values. However, let us turn to the graphical approach and simply work in the
reverse order employed in the preceding sections. The JFET transfer characteristics
are first sketched as shown in Fig. 6.46. The level of IDQ is then established by a horizontal line as shown in the same figure. VGSQ is then determined by dropping a line
down from the operating point to the horizontal axis, resulting in
VGSQ 3.7 V
The level of VC:
VC VB VGSQ 3.62 V (3.7 V)
7.32 V
ID (mA)
12 I DSS
10
8
6
4
2
Q-point
I D = 1.825 mA
Q
– 6 –5
VP
–4
–3
–2
–1
VGS ≅ – 3.7 V
Q
0
Figure 6.46 Determining the
Q-point for the network of Fig.
6.45.
6.8
Combination Networks
283
EXAMPLE 6.14
Determine VD for the network of Fig. 6.47.
Figure 6.47 Example 6.14.
Solution
In this case, there is no obvious path to determine a voltage or current level for the
transistor configuration. However, turning to the self-biased JFET, an equation for
VGS can be derived and the resulting quiescent point determined using graphical
techniques. That is,
VGS IDRS ID(2.4 k)
resulting in the self-bias line appearing in Fig. 6.48 that establishes a quiescent
point at
VGSQ 2.6 V
IDQ 1 mA
ID (mA)
8 IDSS
For the transistor,
IE ≅ IC ID 1 mA
7
6
4
VB 16 V IB(470 k)
3
2
1
16 V (12.5 A)(470 k) 16 V 5.875 V
1.67 mA
I D = 1 mA
10.125 V
Q
0
–4 –3 –2 –1
VP
VGS = –2.6 V
and
VE VD VB VBE
10.125 V 0.7 V
Q
Figure 6.48 Determining the
Q-point for the network of Fig.
6.47.
284
I
1 mA
IB C 12.5 A
80
and
5
9.425 V
Chapter 6
FET Biasing
6.9 DESIGN
The design process is one that is not limited solely to dc conditions. The area of application, level of amplification desired, signal strength, and operating conditions are
just a few of the conditions that enter into the total design process. However, we will
first concentrate on establishing the chosen dc conditions.
For example, if the levels of VD and ID are specified for the network of Fig. 6.49,
the level of VGSQ can be determined from a plot of the transfer curve and RS can then
be determined from VGS IDRS. If VDD is specified, the level of RD can then be
calculated from RD (VDD VD)/ID. Of course, the value of RS and RD may not be
standard commercial values, requiring that the nearest commercial value be employed.
However, with the tolerance (range of values) normally specified for the parameters
of a network, the slight variation due to the choice of standard values will seldom
cause a real concern in the design process.
The above is only one possibility for the design phase involving the network of
Fig. 6.49. It is possible that only VDD and RD are specified together with the level of
VDS. The device to be employed may have to be specified along with the level of RS.
It appears logical that the device chosen should have a maximum VDS greater than
the specified value by a safe margin.
In general, it is good design practice for linear amplifiers to choose operating
points that do not crowd the saturation level (IDSS) or cutoff (VP) regions. Levels of
VGSQ close to VP/2 or IDQ near IDSS/2 are certainly reasonable starting points in the
design. Of course, in every design procedure the maximum levels of ID and VDS as
appearing on the specification sheet must not be considered as exceeded.
The examples to follow have a design or synthesis orientation in that specific levels are provided and network parameters such as RD, RS, VDD, and so on, must be determined. In any case, the approach is in many ways the opposite of that described
in previous sections. In some cases, it is just a matter of applying Ohm’s law in its
appropriate form. In particular, if resistive levels are requested, the result is often obtained simply by applying Ohm’s law in the following form:
VR
Runknown IR
Figure 6.49 Self-bias configuration to be designed.
(6.34)
where VR and IR are often parameters that can be found directly from the specified
voltage and current levels.
For the network of Fig. 6.50, the levels of VDQ and IDQ are specified. Determine the
required values of RD and RS. What are the closest standard commercial values?
EXAMPLE 6.15
20 V
I D = 2.5 mA
Q
RD
VD = 12 V
I DSS = 6 mA
VP = – 3 V
RS
Figure 6.50 Example 6.15.
6.9
Design
285
Solution
As defined by Eq. (6.34),
VDD VDQ
VRD
RD IDQ
IDQ
20 V 12 V
8V
3.2 k
2.5 mA
2.5 mA
and
Plotting the transfer curve in Fig. 6.51 and drawing a horizontal line at IDQ 2.5 mA will result in VGSQ 1 V, and applying VGS IDRS will establish the
level of RS:
(VGSQ )
(1 V)
RS 0.4 k
IDQ
2.5 mA
ID (mA)
6 IDSS
5
4
3
I = 2.5 mA
2 DQ
1
–3 –2
VP
–1
0
VGS = – 1 V
Q
VGS
Figure 6.51 Determining VGSQ
for the network of Fig. 6.50.
The nearest standard commercial values are
RD 3.2 k ⇒ 3.3 k
RS 0.4 k ⇒ 0.39 k
EXAMPLE 6.16
For the voltage-divider bias configuration of Fig. 6.52, if VD 12 V and VGSQ 2 V, determine the value of RS.
Solution
The level of VG is determined as follows:
47 k(16 V)
VG 5.44 V
47 k 91 k
VDD VD
ID RD
with
16 V 12 V
2.22 mA
1.8 k
The equation for VGS is then written and the known values substituted:
VGS VG IDRS
2 V 5.44 V (2.22 mA)RS
Figure 6.52 Example 6.16.
7.44 V (2.22 mA)RS
7.44 V
RS 3.35 k
2.22 mA
and
The nearest standard commercial value is 3.3 k.
286
Chapter 6
FET Biasing
The levels of VDS and ID are specified as VDS 12VDD and ID ID(on) for the network
of Fig. 6.53. Determine the level of VDD and RD.
EXAMPLE 6.17
VDD
RD
10 MΩ
VGS(on) = 6 V
I D(on) = 4 mA
VGS(Th) = 3 V
Figure 6.53 Example 6.17.
Solution
Given ID ID(on) 4 mA and VGS VGS(on) 6 V, for this configuration,
VDS VGS 12VDD
and
6 V 12VDD
so that
VDD 12 V
Applying Eq. (6.34) yields
12V
DD
VDD 21VDD
VRD
VDD VDS RD I
D(on)
I
D(on)
ID
ID(on)
and
6V
RD 1.5 k
4 mA
which is a standard commercial value.
6.10 TROUBLESHOOTING
How often has a network been carefully constructed only to find that when the power
is applied, the response is totally unexpected and fails to match the theoretical calculations. What is the next step? Is it a bad connection? A misreading of the color
code for a resistive element? An error in the construction process? The range of possibilities seems vast and often frustrating. The troubleshooting process first described
in the analysis of BJT transistor configurations should narrow down the list of possibilities and isolate the problem area following a definite plan of attack. In general,
the process begins with a rechecking of the network construction and the terminal
connections. This is usually followed by the checking of voltage levels between specific terminals and ground or between terminals of the network. Seldom are current
levels measured since such maneuvers require disturbing the network structure to insert the meter. Of course, once the voltage levels are obtained, current levels can be
calculated using Ohm’s law. In any case, some idea of the expected voltage or current level must be known for the measurement to have any importance. In total, therefore, the troubleshooting process can begin with some hope of success only if the basic operation of the network is understood along with some expected levels of voltage
6.10
Troubleshooting
287
Figure 6.54 Checking the dc
operation of the JFET self-bias
configuration.
or current. For the n-channel JFET amplifier, it is clearly understood that the quiescent value of VGSQ is limited to 0 V or a negative voltage. For the network of Fig.
6.54, VGSQ is limited to negative values in the range 0 V to VP. If a meter is hooked
up as shown in Fig. 6.54, with the positive lead (normally red) to the gate and the
negative lead (usually black) to the source, the resulting reading should have a negative sign and a magnitude of a few volts. Any other response should be considered
suspicious and needs to be investigated.
The level of VDS is typically between 25% and 75% of VDD. A reading of 0 V for
VDS clearly indicates that either the output circuit has an “open” or the JFET is internally short-circuited between drain and source. If VD is VDD volts, there is obviously no drop across RD due to the lack of current through RD and the connections
should be checked for continuity.
If the level of VDS seems inappropriate, the continuity of the output circuit can
easily be checked by grounding the negative lead of the voltmeter and measuring the
voltage levels from VDD to ground using the positive lead. If VD VDD, the current
through RD may be zero, but there is continuity between VD and VDD. If VS VDD,
the device is not open between drain and source, but it is also not “on.” The continuity through to VS is confirmed, however. In this case, it is possible that there is a
poor ground connection between RS and ground that may not be obvious. The internal connection between the wire of your lead and the terminal connector may have
separated. Other possibilities also exist, such as a shorted device from drain to source,
but the troubleshooter will simply have to narrow down the possible causes for the
malfunction.
The continuity of a network can also be checked simply by measuring the voltage across any resistor of the network (except for RG in the JFET configuration). An
indication of 0 V immediately reveals the lack of current through the element due to
an open circuit in the network.
The most sensitive element in the BJT and JFET configurations is the amplifier
itself. The application of excessive voltage during the construction or testing phase
or the use of incorrect resistor values resulting in high current levels can destroy the
device. If you question the condition of the amplifier, the best test for the FET is the
curve tracer since it not only reveals whether the device is operable but also its range
of current and voltage levels. Some testers may reveal that the device is still fundamentally sound but do not reveal whether its range of operation has been severely reduced.
The development of good troubleshooting techniques comes primarily from experience and a level of confidence in what to expect and why. There are, of course,
times when the reasons for a strange response seem to disappear mysteriously when
you check a network. In such cases, it is best not to breathe a sigh of relief and
continue with the construction. The cause for such a sensitive “make or break”
situation should be found and corrected, or it may reoccur at the most inopportune
moment.
6.11 P-CHANNEL FETS
The analysis thus far has been limited solely to n-channel FETs. For p-channel FETs,
a mirror image of the transfer curves is employed, and the defined current directions
are reversed as shown in Fig. 6.55 for the various types of FETs.
Note for each configuration of Fig. 6.55 that each supply voltage is now a negative voltage drawing current in the indicated direction. In particular, note that the
double-subscript notation for voltages continues as defined for the n-channel device:
VGS, VDS, and so on. In this case, however, VGS is positive (positive or negative for
the depletion-type MOSFET) and VDS negative.
288
Chapter 6
FET Biasing
Figure 6.55 p-channel configurations.
Due to the similarities between the analysis of n-channel and p-channel devices,
one can actually assume an n-channel device and reverse the supply voltage and perform the entire analysis. When the results are obtained, the magnitude of each quantity will be correct, although the current direction and voltage polarities will have to
be reversed. However, the next example will demonstrate that with the experience
gained through the analysis of n-channel devices, the analysis of p-channel devices
is quite straightforward.
6.11 P-Channel FETs
289
EXAMPLE 6.18
Determine IDQ, VGSQ, and VDS for the p-channel JFET of Fig. 6.56.
Figure 6.56 Example 6.18.
Solution
20 k(20 V)
VG 4.55 V
20 k 68 k
Applying Kirchhoff’s voltage law gives
VG VGS IDRS 0
VGS VG IDRS
and
Choosing ID 0 mA yields
VGS VG 4.55 V
as appearing in Fig. 6.57.
Choosing VGS 0 V, we obtain
VG
4.55 V
ID 2.53 mA
RS
1.8 k
as also appearing in Fig. 6.57.
The resulting quiescent point from Fig. 6.57:
IDQ 3.4 mA
VGSQ 1.4 V
ID (mA)
8
7
6
5
I D = 3.4 mA
4
Q- point
Q
2
1
– 5 – 4 –3 – 2 – 1
0 1
2
VGS
4
VP
= 1.4 V
Q
290
Chapter 6
FET Biasing
3
VGS
Figure 6.57 Determining the
Q-point for the JFET configuration of Fig. 6.56.
For VDS, Kirchhoff’s voltage law will result in
IDRS VDS IDRD VDD 0
VDS VDD ID(RD RS)
and
20 V (3.4 mA)(2.7 k 1.8 k)
20 V 15.3 V
4.7 V
6.12 UNIVERSAL JFET BIAS CURVE
Since the dc solution of a FET configuration requires drawing the transfer curve for
each analysis, a universal curve was developed that can be used for any level of IDSS
and VP. The universal curve for an n-channel JFET or depletion-type MOSFET (for
negative values of VGSQ) is provided in Fig. 6.58. Note that the horizontal axis is not
that of VGS but of a normalized level defined by VGS /VP, the VP indicating that
only the magnitude of VP is to be employed, not its sign. For the vertical axis, the
scale is also a normalized level of ID /IDSS. The result is that when ID IDSS, the
ratio is 1, and when VGS VP, the ratio VGS /VPis 1. Note also that the scale
for ID/IDSS is on the left rather than on the right as encountered for ID in past exercises. The additional two scales on the right need an introduction. The vertical scale
labeled m can in itself be used to find the solution to fixed-bias configurations.
The other scale, labeled M, is employed along with the m scale to find the solution
m=
VP
RS IDSS
M= m
+
ID
I DSS
1.0
5
1.0
0.8
4
0.8
0.6
3
0.6
0.4
2
0.4
0.2
1
0.2
Normalized curve
V
of ID = I DSS 1 – GS
VP
0
–1
– 0.8
VG G
VP
2
– 0.6
– 0.4
– 0.2
6.12
0
VGS
VP
Figure 6.58 Universal JFET bias
curve.
Universal JFET Bias Curve
291
to voltage-divider configurations. The scaling for m and M come from a mathematical development involving the network equations and normalized scaling just introduced. The description to follow will not concentrate on why the m scale extends from
0 to 5 at VGS /VP 0.2 and the M scale from 0 to 1 at VGS /VP 0 but rather
on how to use the resulting scales to obtain a solution for the configurations. The
equations for m and M are the following, with VG as defind by Eq. (6.15).
VP
m IDSSRS
Mm
(6.35)
VG
VP
(6.36)
R2VDD
VG R1 R2
with
Keep in mind that the beauty of this approach is the elimination of the need to sketch
the transfer curve for each analysis, that the superposition of the bias line is a great
deal easier, and that the calculations are fewer. The use of the m and M axes is best
described by examples employing the scales. Once the procedure is clearly understood, the analysis can be quite rapid, with a good measure of accuracy.
EXAMPLE 6.19
Determine the quiescent values of ID and VGS for the network of Fig. 6.59.
Figure 6.59 Example 6.19.
Solution
Calculating the value of m, we obtain
VP
 3 V
m 0.31
IDSSRS
(6 mA)(1.6 k)
The self-bias line defined by RS is plotted by drawing a straight line from the origin
through a point defined by m 0.31, as shown in Fig. 6.60.
The resulting Q-point:
VGS
I
D 0.18
0.575
and
VP
IDSS
292
Chapter 6
FET Biasing
Figure 6.60 Universal curve for Examples 6.19 and 6.20.
The quiescent values of ID and VGS can then be determined as follows:
IDQ 0.18IDSS 0.18(6 mA) 1.08 mA
and
VGSQ 0.575VP 0.575(3 V) 1.73 V
Determine the quiescent values of ID and VGS for the network of Fig. 6.61.
Figure 6.61 Example 6.20.
EXAMPLE 6.20
293
Solution
Calculating m gives
VP
6 V
m 0.625
IDSSRS
(8 mA)(1.2 k)
Determining VG yields
R2VDD
(220 k)(18 V)
VG 3.5 V
R1 R2
910 k 220 k
Finding M, we have
Mm
VG
3.5 V
0.625 0.365
VP
6V
Now that m and M are known, the bias line can be drawn on Fig. 6.60. In particular,
note that even though the levels of IDSS and VP are different for the two networks, the
same universal curve can be employed. First find M on the M axis as shown in Fig.
6.60. Then draw a horizontal line over to the m axis and, at the point of intersection,
add the magnitude of m as shown in the figure. Using the resulting point on the m
axis and the M intersection, draw the straight line to intersect with the transfer curve
and define the Q-point:
VGS
I
D 0.53
0.26
and
That is,
VP
IDSS
and
IDQ 0.53IDSS 0.53(8 mA) 4.24 mA
with
VGSQ 0.26VP 0.26(6 V) 1.56 V
6.13 PSPICE WINDOWS
JFET Voltage-Divider Configuration
The results of Example 6.20 will now be verified using PSpice Windows. The network of Fig. 6.62 is constructed using computer methods described in the previous
chapters. The J2N3819 JFET is obtained from the EVAL.slb library and, through
Edit-Model-Edit Instance Model (Text), Vto is set to 6V and Beta, as defined by
Beta IDSS/VP2 is set to 0.222 mA/V2. After an OK followed by clicking the
Simulation icon (the yellow background with the two waveforms) and clearing the
Message Viewer, PSpiceAD screens will result in Fig. 6.62. The resulting drain cur-
Figure 6.62 JFET voltage-divider configuration with PSpice Windows results
for the dc levels.
294
Chapter 6
FET Biasing
rent is 4.231 mA compared to the calculated level of 4.24 mA, and VGS is 3.504 V 5.077 V 1.573 V versus the calculated value of 1.56 V—both excellent comparisons.
Combination Network
Next, the results of Example 6.13 with both a transistor and JFET will be verified.
For the transistor, the Model must be altered to have a Bf(beta) of 180 to match the
example, and for the JFET, Vto must be set to 6V and Beta to 0.333 mA/V2. The
results appearing in Fig. 6.63 are again an excellent comparison with the handwritten solution. VD is 11.44 V compared to 11.07 V, VC is 7.138 V compared to 7.32 V,
and VGS is 3.758 V compared to 3.7 V.
Figure 6.63 Verifying
the hand-calculated solution of Example 6.13 using PSpice Windows.
Enhancement MOSFET
Next, the analysis procedure of Section 6.6 will be verified using the IRF150
enhancement-type n-channel MOSFET found in the EVAL.slb library. First, the device characteristics will be obtained by constructing the network of Fig. 6.64.
Figure 6.64 Network employed to obtain the characteristics of the IRF150 enhancement-type n-channel
MOSFET.
Clicking on the Setup Analysis icon (with the blue bar at the top in the left-hand
corner of the screen), DC Sweep is chosen to obtain the DC Sweep dialog box.
Voltage Source is chosen as the Swept Var. Type, and Linear is chosen for the Sweep
Type. Since only one curve will be obtained, there is no need for a Nested Sweep.
The voltage-drain voltage VDD will remain fixed at a value of 9 V (about three times
the threshold value (Vto) of 2.831 V), while the gate-to-source voltage VGS, which in
6.13
PSpice Windows
295
this case is VGG, will be swept from 0 to 10 V. The Name therefore is VGG and the
Start Value 0V, the End Value 10V, and the Increment 0.01V. After an OK followed
by a Close of the Analysis Setup, the analysis can be performed through the Analysis icon. If Automatically run Probe after simulation is chosen under the Probe
Setup Options of Analysis, the OrCAD-MicroSim Probe screen will result, with
the horizontal axis appearing with VGG as the variable and range from 0 to 10 V.
Next, the Add Traces dialog box can be obtained by clicking the Traces icon (red
pointed pattern on an axis) and the ID(M1) chosen to obtain the drain current versus
the gate-to-source voltage. Click OK, and the characteristics will appear on the screen.
To expand the scale of the resulting plot to 20 V, simply choose Plot followed by XAxis Settings and set the User Defined range to 0 to 20 V. After another OK, and
the plot of Fig. 6.65 will result, revealing a rather high-current device. The labels ID
and VGS were added using the Text Label icon with the letters A, B, and C. The
hand-drawn load line will be described in the paragraph to follow.
Figure 6.65 Characteristics of the IRF500 MOSFET of Figure 6.64 with a load line defined by the
network of Figure 6.66.
The network of Fig. 6.66 was then established to provide a load line extending
from ID equal to 20 V/0.4 Ω 50 A down to VGS VGG 20 V as shown in Fig.
6.65. A simulation resulted in the levels shown, which match the solution of Fig. 6.65.
Figure 6.66 Feedback-biasing arrangement employing an IRF150 enhancement-type MOSFET.
296
Chapter 6
FET Biasing
§ 6.2 Fixed-Bias Configuration
PROBLEMS
1. For the fixed-bias configuration of Fig. 6.67:
(a) Sketch the transfer characteristics of the device.
(b) Superimpose the network equation on the same graph.
(c) Determine IDQ and VDSQ.
(d) Using Shockley’s equation, solve for IDQ and then find VDSQ. Compare with the solutions
of part (c).
Figure 6.67 Problems 1, 35
2. For the fixed-bias configuration of Fig. 6.68, determine:
(a) IDQ and VGSQ using a purely mathematical approach.
(b) Repeat part (a) using a graphical approach and compare results.
(c) Find VDS, VD, VG, and VS using the results of part (a).
Figure 6.68 Problem 2
3. Given the measured value of VD in Fig. 6.69, determine:
(a) ID.
(b) VDS.
(c) VGG.
Figure 6.69 Problem 3
297
4. Determine VD for the fixed-bias configuration of Fig. 6.70.
5. Determine VD for the fixed-bias configuration of Fig. 6.71.
Figure 6.70 Problem 4
Figure 6.71 Problem 5
§ 6.3 Self-Bias Configuration
6.
For the self-bias configuration of Fig. 6.72:
(a) Sketch the transfer curve for the device.
(b) Superimpose the network equation on the same graph.
(c) Determine IDQ and VGSQ.
(d) Calculate VDS, VD, VG, and VS.
Figure 6.72 Problems 6, 7, 36
* 7. Determine IDQ for the network of Fig. 6.72 using a purely mathematical approach. That is, establish a quadratic equation for ID and choose the solution compatible with the network characteristics. Compare to the solution obtained in Problem 6.
298
8.
For the network of Fig. 6.73, determine:
(a) VGSQ and IDQ.
(b) VDS, VD, VG, and VS.
9.
Given the measurement VS 1.7 V for the network of Fig. 6.74, determine:
(a) IDQ.
(b) VGSQ.
(c) IDSS.
(d) VD.
(e) VDS.
Chapter 6
FET Biasing
* 10. For the network of Fig. 6.75, determine:
(a) ID.
(b) VDS.
(c) VD.
(d) VS.
Figure 6.73 Problem 8
Figure 6.74 Problem 9
Figure 6.75 Problem 10
* 11. Find VS for the network of Fig. 6.76.
Figure 6.76 Problem 11
§ 6.4 Voltage-Divider Biasing
12. For the network of Fig. 6.77, determine:
(a) VG.
(b) IDQ and VGSQ.
(c) VD and VS.
(d) VDSQ.
Figure 6.77 Problems 12, 13
13. (a) Repeat Problem 12 with RS 0.51 k (about 50% of the value of 12). What is the effect
of a smaller RS on IDQ and VGSQ?
(b) What is the minimum possible value of RS for the network of Fig. 6.77?
Problems
299
14.
For the network of Fig. 6.78, VD 9 V. Determine:
(a) ID.
(b) VS and VDS.
(c) VG and VGS.
(d) VP.
* 15. For the network of Fig. 6.79, determine:
(a) IDQ and VGSQ.
(b) VDS and VS.
18 V
ID
2 kΩ
750 kΩ
VG
+
VGS
–
VS
+
VD = 9 V
VDS
IDSS = 8 mA
–
91 kΩ
0.68 kΩ
Figure 6.79 Problems 15, 37
Figure 6.78 Problem 14
* 16. Given VDS 4 V for the network of Fig. 6.80, determine:
(a) ID.
(b) VD and VS.
(c) VGS.
§ 6.5 Depletion-Type MOSFETs
17.
For the self-bias configuration of Fig. 6.81, determine:
(a) IDQ and VGSQ.
(b) VDS and VD.
* 18. For the network of Fig. 6.82, determine:
(a) IDQ and VGSQ.
(b) VDS and VS.
Figure 6.80 Problem 16
Figure 6.82 Problem 18
Figure 6.81 Problem 17
300
Chapter 6
FET Biasing
§ 6.6 Enhancement-Type MOSFETs
19.
For the network of Fig. 6.83, determine:
(a) IDQ.
(b) VGSQ and VDSQ.
(c) VD and VS.
(d) VDS.
20.
For the voltage-divider configuration of Fig. 6.84, determine:
(a) IDQ and VGSQ.
(b) VD and VS.
24 V
2.2 kΩ
10 MΩ
ID
Q
+
VGS(Th) = 3 V
I D(on) = 5 mA
VGS(on) = 6 V
VGS
Q
–
6.8 MΩ
0.75 kΩ
Figure 6.83 Problem 19
Figure 6.84 Problem 20
§ 6.8 Combination Networks
*
21. For the network of Fig. 6.85, determine:
(a) VG.
(b) VGSQ and IDQ.
(c) IE.
(d) IB.
(e) VD.
(f) VC.
Figure 6.85 Problem 21
Problems
301
* 22. For the combination network of Fig. 6.86, determine:
(a) VB and VG.
(b) VE.
(c) IE, IC, and ID.
(d) IB.
(e) VC, VS, and VD.
(f) VCE.
(g) VDS.
Figure 6.86 Problem 22
§ 6.9 Design
* 23. Design a self-bias network using a JFET transistor with IDSS 8 mA and VP 6 V to have a
Q-point at IDQ 4 mA using a supply of 14 V. Assume that RD 3RS and use standard values.
* 24. Design a voltage-divider bias network using a depletion-type MOSFET with IDSS 10 mA and
VP 4 V to have a Q-point at IDQ 2.5 mA using a supply of 24 V. In addition, set VG 4
V and use RD 2.5RS with R1 22 M. Use standard values.
25. Design a network such as appears in Fig. 6.39 using an enhancement-type MOSFET with
VGS(Th) 4 V, k 0.5 103A/V2 to have a Q-point of IDQ 6 mA. Use a supply of 16 V
and standard values.
§ 6.10 Troubleshooting
* 26. What do the readings for each configuration of Fig. 6.87 suggest about the operation of the
network?
Figure 6.87 Problem 26
302
Chapter 6
FET Biasing
*
27. Although the readings of Fig. 6.88 initially suggest that the network is behaving properly, determine a possible cause for the undesirable state of the network.
*
28. The network of Fig. 6.89 is not operating properly. What is the specific cause for its failure?
Figure 6.88 Problem 27
Figure 6.89 Problem 28
§ 6.11 p-Channel FETs
29. For the network of Fig. 6.90, determine:
(a) IDQ and VGSQ.
(b) VDS.
(c) VD.
30. For the network of Fig. 6.91, determine:
(a) IDQ and VGSQ.
(b) VDS.
(c) VD.
Figure 6.90 Problem 29
Figure 6.91 Problem 30
§ 6.12 Universal JFET Bias Curve
31. Repeat Problem 1 using the universal JFET bias curve.
32. Repeat Problem 6 using the universal JFET bias curve.
33. Repeat Problem 12 using the universal JFET bias curve.
34. Repeat Problem 15 using the universal JFET bias curve.
Problems
303
§ 6.13 PSpice Windows
35. Perform a PSpice Windows analysis of the network of Problem 1.
36. Perform a PSpice Windows analysis of the network of Problem 6.
37. Perform a PSpice Windows analysis of the network of Problem 15.
*Please Note: Asterisks indicate more difficult problems.
304
Chapter 6
FET Biasing
re
CHAPTER
BJT Transistor
Modeling
7
7.1 INTRODUCTION
The basic construction, appearance, and characteristics of the transistor were introduced in Chapter 3. The dc biasing of the device was then examined in detail in Chapter 4. We now begin to examine the small-signal ac response of the BJT amplifier by
reviewing the models most frequently used to represent the transistor in the sinusoidal
ac domain.
One of our first concerns in the sinusoidal ac analysis of transistor networks
is the magnitude of the input signal. It will determine whether small-signal or largesignal techniques should be applied. There is no set dividing line between the two,
but the application––and the magnitude of the variables of interest relative to the
scales of the device characteristics––will usually make it quite clear which method is
appropriate. The small-signal technique is introduced in this chapter, and large-signal
applications are examined in Chapter 16.
There are two models commonly used in the small-signal ac analysis of transistor networks: the re model and the hybrid equivalent model. This chapter not only introduces both models but defines the role of each and the relationship between the
two.
7.2 AMPLIFICATION IN THE AC DOMAIN
It was demonstrated in Chapter 3 that the transistor can be employed as an amplifying device. That is, the output sinusoidal signal is greater than the input signal or,
stated another way, the output ac power is greater than the input ac power. The question then arises as to how the ac power output can be greater than the input ac power?
Conservation of energy dictates that over time the total power output, Po, of a system
cannot be greater than its power input, Pi, and that the efficiency defined by Po /Pi
cannot be greater than 1. The factor missing from the discussion above that permits
an ac power output greater than the input ac power is the applied dc power. It is a
contributor to the total output power even though part of it is dissipated by the device and resistive elements. In other words, there is an “exchange” of dc power to the
ac domain that permits establishing a higher output ac power. In fact, a conversion
efficiency is defined by Po(ac)/Pi(dc), where Po(ac) is the ac power to the load and
Pi(dc) is the dc power supplied.
Perhaps the role of the dc supply can best be described by first considering the
simple dc network of Fig. 7.1. The resulting direction of flow is indicated in the figure with a plot of the current i versus time. Let us now insert a control mechanism
Figure 7.1 Steady current
established by a dc supply.
305
re
such as that shown in Fig. 7.2. The control mechanism is such that the application of
a relatively small signal to the control mechanism can result in a much larger oscillation in the output circuit. For the system of Fig. 7.2, the peak value of the oscillation is controlled by the established dc level. Any attempt to exceed the limit set by
the dc level will result in a “clipping” (flattening) of the peak region of the output
signal. In total, therefore, proper amplifier design requires that the dc and ac components be sensitive to each other’s requirements and limitations.
However, it is indeed fortunate that transistor small-signal amplifiers can be
considered linear for most applications, permitting the use of the superposition theorem to isolate the dc analysis from the ac analysis.
7.3 BJT TRANSISTOR MODELING
Figure 7.2 Effect of a control
element on the steady-state flow
of the electrical system of Fig. 7.1.
The key to transistor small-signal analysis is the use of equivalent circuits (models)
to be introduced in this chapter.
A model is the combination of circuit elements, properly chosen, that best approximates the actual behavior of a semiconductor device under specific operating conditions.
Once the ac equivalent circuit has been determined, the graphical symbol of the
device can be replaced in the schematic by this circuit and the basic methods of ac
circuit analysis (mesh analysis, nodal analysis, and Thévenin’s theorem) can be applied to determine the response of the circuit.
There are two schools of thought in prominence today regarding the equivalent
circuit to be substituted for the transistor. For many years the industrial and educational institutions relied heavily on the hybrid parameters (to be introduced shortly).
The hybrid-parameter equivalent circuit continues to be very popular, although it must
now share the spotlight with an equivalent circuit derived directly from the operating
conditions of the transistor—the re model. Manufacturers continue to specify the hybrid parameters for a particular operating region on their specification sheets. The parameters (or components) of the re model can be derived directly from the hybrid parameters in this region. However, the hybrid equivalent circuit suffers from being
limited to a particular set of operating conditions if it is to be considered accurate.
The parameters of the other equivalent circuit can be determined for any region of
operation within the active region and are not limited by the single set of parameters
provided by the specification sheet. In turn, however, the re model fails to account for
the output impedance level of the device and the feedback effect from output to
input.
Since both models are used extensively today, they are both examined in detail in
this text. In some analysis and examples the hybrid model will be employed, while
in others the re model will be used exclusively. The text will make every effort, however, to show how closely related the two models are and how a proficiency with one
leads to a natural proficiency with the other.
In an effort to demonstrate the effect that the ac equivalent circuit will have on
the analysis to follow, consider the circuit of Fig. 7.3. Let us assume for the moment
that the small-signal ac equivalent circuit for the transistor has already been determined. Since we are interested only in the ac response of the circuit, all the dc supplies can be replaced by a zero-potential equivalent (short circuit) since they determine only the dc (quiescent level) of the output voltage and not the magnitude of the
swing of the ac output. This is clearly demonstrated by Fig. 7.4. The dc levels were
simply important for determining the proper Q-point of operation. Once determined,
the dc levels can be ignored in the ac analysis of the network. In addition, the coupling capacitors C1 and C2 and bypass capacitor C3 were chosen to have a very small
306
Chapter 7
BJT Transistor Modeling
re
Figure 7.3 Transistor circuit
under examination in this introductory discussion.
Figure 7.4 The network of
Fig. 7.3 following removal of the
dc supply and insertion of the
short-circuit equivalent for the capacitors.
reactance at the frequency of application. Therefore, they too may for all practical
purposes be replaced by a low-resistance path or a short circuit. Note that this will
result in the “shorting out” of the dc biasing resistor RE. Recall that capacitors assume an “open-circuit” equivalent under dc steady-state conditions, permitting an isolation between stages for the dc levels and quiescent conditions.
If we establish a common ground and rearrange the elements of Fig. 7.4, R1 and
R2 will be in parallel and RC will appear from collector to emitter as shown in Fig.
7.5. Since the components of the transistor equivalent circuit appearing in Fig. 7.5
employ familiar components such as resistors and independent controlled sources,
Figure 7.5 Circuit of Fig. 7.4 redrawn for small-signal ac analysis.
7.3 BJT Transistor Modeling
307
re
analysis techniques such as superposition, Thévenin’s theorem, and so on, can be applied to determine the desired quantities.
Let us further examine Fig. 7.5 and identify the important quantities to be determined for the system. Since we know that the transistor is an amplifying device, we
would expect some indication of how the output voltage Vo is related to the input
voltage Vi —the voltage gain. Note in Fig. 7.5 for this configuration that Ii Ib and
Io Ic, which define the current gain Ai Io/Ii. The input impedance Zi and output
impedance Zo will prove particularly important in the analysis to follow. A great deal
more will be offered about these parameters in the sections to follow.
In summary, therefore, the ac equivalent of a network is obtained by:
1. Setting all dc sources to zero and replacing them by a short-circuit equivalent
2. Replacing all capacitors by a short-circuit equivalent
3. Removing all elements bypassed by the short-circuit equivalents introduced
by steps 1 and 2
4. Redrawing the network in a more convenient and logical form
In the sections to follow, the re and hybrid equivalent circuits will be introduced
to complete the ac analysis of the network of Fig. 7.5.
7.4 THE IMPORTANT PARAMETERS:
Zi, Zo, Av, Ai
Before investigating the equivalent circuits for BJTs in some detail, let us concentrate
on those parameters of a two-port system that are of paramount importance from an
analysis and design viewpoint. For the two-port (two pairs of terminals) system of
Fig. 7.6, the input side (the side to which the signal is normally applied) is to the left
and the output side (where the load is connected) is to the right. In fact, for most electrical and electronic systems, the general flow is usually from the left to the right. For
both sets of terminals, the impedance between each pair of terminals under normal
operating conditions is quite important.
Figure 7.6
Two-port system.
Input Impedance, Zi
For the input side, the input impedance Zi is defined by Ohm’s law as the following:
V
Zi i
Ii
(7.1)
If the input signal Vi is changed, the current Ii can be computed using the same
level of input impedance. In other words:
308
Chapter 7
BJT Transistor Modeling
re
For small-signal analysis, once the input impedance has been determined the
same numerical value can be used for changing levels of applied signal.
In fact, we will find in the sections to follow that the input impedance of a transistor can be approximately determined by the dc biasing conditions—conditions that
do not change simply because the magnitude of the applied ac signal has changed.
It is particularly noteworthy that for frequencies in the low to mid-range (typically 100 kHz):
The input impedance of a BJT transistor amplifier is purely resistive in nature
and, depending on the manner in which the transistor is employed, can vary
from a few ohms to megohms.
In addition:
An ohmmeter cannot be used to measure the small-signal ac input impedance
since the ohmmeter operates in the dc mode.
Equation (7.1) is particularly useful in that it provides a method for measuring
the input resistance in the ac domain. For instance, in Fig. 7.7 a sensing resistor has
been added to the input side to permit a determination of Ii using Ohm’s law. An oscilloscope or sensitive digital multimeter (DMM) can be used to measure the voltage
Vs and Vi. Both voltages can be the peak-to-peak, peak, or rms values, as long as both
levels use the same standard. The input impedance is then determined in the following manner:
Vs Vi
Ii Rsense
(7.2)
V
Zi i
Ii
(7.3)
and
+
Ii
+
Rsense
Zi
Vs
Vi
Two-port
System
–
–
Figure 7.7
Determining Zi.
The importance of the input impedance of a system can best be demonstrated by
the network of Fig. 7.8. The signal source has an internal resistance of 600 , and
the system (possibly a transistor amplifier) has an input resistance of 1.2 k. If the
source were ideal (Rs 0 ), the full 10 mV would be applied to the system, but
Rsource
+
Vs
600 Ω
10 mV
–
+
Zi = 1.2 kΩ
Vi
–
Amplifier
Figure 7.8 Demonstrating the
impact of Zi on an amplifier’s response.
7.4 The Important Parameters: Zi, Zo, Av, Ai
309
re
with a source impedance, the input voltage must be determined using the voltagedivider rule as follows:
Z Vs
(1.2 k)(10 mV)
6.67 mV
Vi i Zi Rsource
1.2 k 0.6 k
Thus, only 66.7% of the full-input signal is available at the input. If Zi were only
600 , then Vi 12(10 mV) 5 mV or 50% of the available signal. Of course, if
Zi 8.2 k, Vi will be 93.2% of the applied signal. The level of input impedance,
therefore, can have a significant impact on the level of signal that reaches the system
(or amplifier). In the sections and chapters to follow, it will be demonstrated that the
ac input resistance is dependent on whether the transistor is in the common-base,
common-emitter, or common-collector configuration and on the placement of the resistive elements.
EXAMPLE 7.1
For the system of Fig. 7.9, determine the level of input impedance.
Rsense
1 kΩ
+
Vs
2 mV
Zi
+
Vi = 1.2 mV
–
Two-port
System
–
Figure 7.9
Example 7.1
Solution
Vs Vi
2 mV 1.2 mV
0.8 mV
Ii 0.8 A
Rsense
1 k
1 k
and
V
1.2 mV
Zi i 1.5 k
Ii
0.8 A
Output Impedance, Zo
The output impedance is naturally defined at the output set of terminals, but the manner in which it is defined is quite different from that of the input impedance. That is:
The output impedance is determined at the output terminals looking back into
the system with the applied signal set to zero.
In Fig. 7.10, for example, the applied signal has been set to zero volts. To determine Zo, a signal, Vs, is applied to the output terminals and the level of Vo is measured with an oscilloscope or sensitive DMM. The output impedance is then
determined in the following manner:
and
310
Chapter 7
BJT Transistor Modeling
VV
Io o
Rsense
(7.4)
Vo
Zo Io
(7.5)
re
Rsource
Rsense
+
Io
Two-port
System
Vs = 0 V
Vo
+
V
Zo
–
–
Figure 7.10 Determining Zo.
In particular, for frequencies in the low to mid-range (typically 100 kHz):
The output impedance of a BJT transistor amplifier is resistive in nature and,
depending on the configuration and the placement of the resistive elements,
Zo, can vary from a few ohms to a level that can exceed 2 M.
In addition:
An ohmmeter cannot be used to measure the small-signal ac output impedance since the ohmmeter operates in the dc mode.
For amplifier configurations where significant gain in current is desired, the level
of Zo should be as large as possible. As demonstrated by Fig. 7.11, if Zo RL, the
majority of the amplifier output current will pass on to the load. It will be demonstrated in the sections and chapters to follow that Zo is frequently so large compared
to RL that it can be replaced by an open-circuit equivalent.
Figure 7.11 Effect of Zo Ro
on the load or output current IL.
EXAMPLE 7.2
For the system of Fig. 7.12, determine the level of output impedance.
Rsense
Two-port
System
Vs = 0 V
+
20 kΩ
Zo
Vo = 680 mV
–
+
V= 1 V
–
Figure 7.12 Example 7.2.
Solution
VV
1 V 680 mV
320 mV
Io o 16 A
Rsense
20 k
20 k
and
Vo
680 mV
Zo 42.5 k
Io
16 A
Voltage Gain, Av
One of the most important characteristics of an amplifier is the small-signal ac voltage gain as determined by
Vo
Av Vi
(7.6)
7.4 The Important Parameters: Zi, Zo, Av, Ai
311
re
For the system of Fig. 7.13, a load has not been connected to the output terminals
and the level of gain determined by Eq. (7.6) is referred to as the no-load voltage
gain. That is,
Vo
AvNL Vi
(7.7)
RL (open circuit)
Rsource
+
+
Zi
Vs
–
+
AυNL
Vi
–
Vo
–
Figure 7.13 Determining the no-load voltage gain.
In Chapter 9 it will be demonstrated that:
For transistor amplifiers, the no-load voltage gain is greater than the loaded
voltage gain.
For the system of Fig. 7.13 having a source resistance Rs, the level of Vi would
first have to be determined using the voltage-divider rule before the gain Vo/Vs could
be calculated. That is,
Z Vs
Vi i Zi Rs
with
V
Zi
i Vs
Zi Rs
and
V Vo
Vo
Avs i Vs Vi
Vs
so that
Vo
Zi
Avs A
Vs
Zi Rs vNL
(7.8)
Experimentally, the voltage gain Avs or AvNL can be determined simply by measuring the appropriate voltage levels with an oscilloscope or sensitive DMM and substituting into the appropriate equation.
Depending on the configuration, the magnitude of the voltage gain for a
loaded single-stage transistor amplifier typically ranges from just less than 1
to a few hundred. A multistage (multiunit) system, however, can have a voltage gain in the thousands.
EXAMPLE 7.3
312
For the BJT amplifier of Fig. 7.14, determine:
(a) Vi.
(b) Ii.
(c) Zi.
(d) Avs.
Chapter 7
BJT Transistor Modeling
re
Figure 7.14 Example 7.3.
Solution
Vo
Vo
7.68 V
(a) AvNL and Vi 24 mV
AvNL
Vi
320
Vs Vi
40 mV 24 mV
13.33 A
(b) Ii Rs
1.2 k
V
24 mV
(c) Zi i 1.8 k
Ii
13.33 A
Zi
A
(d) Avs Zi Rs vNL
1.8 k
(320)
1.8 k 1.2 k
192
Current Gain, Ai
The last numerical characteristic to be discussed is the current gain defined by
I
Ai o
Ii
(7.9)
Although typically the recipient of less attention than the voltage gain, it is, however,
an important quantity that can have significant impact on the overall effectiveness of
a design. In general:
For BJT amplifiers, the current gain typically ranges from a level just less
than 1 to a level that may exceed 100.
For the loaded situation of Fig. 7.15,
V
and
Ii i
Zi
Vo
Io RL
Figure 7.15 Determining the
loaded current gain.
7.4 The Important Parameters: Zi, Zo, Av, Ai
313
re
with
I
Vo /RL
V Z
Ai o oi
Ii
Vi /Zi
Vi RL
and
Z
Ai Avi
RL
(7.10)
Eq. (7.10) allows the determination of the current gain from the voltage gain and the
impedance levels.
Phase Relationship
The phase relationship between input and output sinusoidal signals is important for
a variety of practical reasons. Fortunately, however:
For the typical transistor amplifier at frequencies that permit ignoring the effects of the reactive elements, the input and output signals are either 180° out
of phase or in phase.
The reason for the either–or situation will become quite clear in the chapters to
follow.
Summary
The parameters of primary importance for an amplifier have now been introduced:
the input impedance Zi, the output impedance Zo, the voltage gain Av, the current gain
Ai, and the resulting phase relationship. Other factors, such as the applied frequency
at the low and high ends of the frequency spectrum, will affect some of these parameters, but this will be discussed in Chapter 11. In the sections and chapters to follow, all the parameters will be determined for a variety of transistor networks to permit a comparison of the strengths and weaknesses for each configuration.
7.5 THE re TRANSISTOR MODEL
The re model employs a diode and controlled current source to duplicate the behavior of a transistor in the region of interest. Recall that a current-controlled current
source is one where the parameters of the current source are controlled by a current
elsewhere in the network. In fact, in general:
BJT transistor amplifiers are referred to as current-controlled devices.
Common Base Configuration
In Fig. 7.16a, a common-base pnp transistor has been inserted within the two-port
structure employed in our discussion of the last few sections. In Fig. 7.16b, the re
model for the transistor has been placed between the same four terminals. As noted
in Section 7.3, the model (equivalent circuit) is chosen in such a way as to approximate the behavior of the device it is replacing in the operating region of interest. In
other words, the results obtained with the model in place should be relatively close
to those obtained with the actual transistor. You will recall from Chapter 3 that one
junction of an operating transistor is forward-biased while the other is reverse-biased.
The forward-biased junction will behave much like a diode (ignoring the effects of
changing levels of VCE) as verified by the curves of Fig. 3.7. For the base-to-emitter
junction of the transistor of Fig. 7.16a, the diode equivalence of Fig. 7.16b between
the same two terminals seems to be quite appropriate. For the output side, recall that
the horizontal curves of Fig. 3.8 revealed that Ic Ie (as derived from Ic Ie) for
the range of values of VCE. The current source of Fig. 7.16b establishes the fact that
314
Chapter 7
BJT Transistor Modeling
re
Figure 7.16 (a) Common-base BJT transistor; (b) re model for the configuration
of Fig. 7.16a.
Ic Ie, with the controlling current Ie appearing in the input side of the equivalent
circuit as dictated by Fig. 7.16a. We have therefore established an equivalence at the
input and output terminals with the current-controlled source, providing a link between the two—an initial review would suggest that the model of Fig. 7.16b is a valid
model of the actual device.
Recall from Chapter 1 that the ac resistance of a diode can be determined by the
equation rac 26 mV/ID, where ID is the dc current through the diode at the Q (quiescent) point. This same equation can be used to find the ac resistance of the diode
of Fig. 7.16b if we simply substitute the emitter current as follows:
26 mV
re IE
(7.11)
The subscript e of re was chosen to emphasize that it is the dc level of emitter
current that determines the ac level of the resistance of the diode of Fig. 7.16b. Substituting the resulting value of re in Fig. 7.16b will result in the very useful model of
Fig. 7.17.
Figure 7.17 Common-base re
equivalent circuit.
Due to the isolation that exists between input and output circuits of Fig. 7.17, it
should be fairly obvious that the input impedance Zi for the common-base configuration of a transistor is simply re. That is,
Zi re
(7.12)
CB
For the common-base configuration, typical values of Zi range from a few
ohms to a maximum of about 50 .
For the output impedance, if we set the signal to zero, then Ie 0 A and Ic Ie (0 A) 0 A, resulting in an open-circuit equivalence at the output terminals.
The result is that for the model of Fig. 7.17,
Zo (7.13)
CB
7.5 The re Transistor Model
315
re
In actuality:
For the common-base configuration, typical values of Zo are in the megohm
range.
The output resistance of the common-base configuration is determined by the slope
of the characteristic lines of the output characteristics as shown in Fig. 7.18. Assuming the lines to be perfectly horizontal (an excellent approximation) would result in
the conclusion of Eq. (7.13). If care were taken to measure Zo graphically or experimentally, levels typically in the range 1- to 2-M would be obtained.
IC (mA)
Slope =
1
ro
IE = 4 mA
4
IE = 3 mA
3
IE = 2 mA
2
IE = 1 mA
1
IE = 0 mA
0
VCB
Figure 7.18 Defining Zo.
In general, for the common-base configuration the input impedance is relatively small and the output impedance quite high.
The voltage gain will now be determined for the network of Fig. 7.19.
Vo Io RL (Ic)RL IeRL
Vi IeZi Iere
and
Vo
I RL
Av e
Vi
Iere
so that
RL
RL
Av re
re
and
(7.14)
CB
For the current gain,
I
I
I
Ai o c e
Ii
Ie
Ie
and
Ai 1
(7.15)
CB
Figure 7.19 Defining Av Vo /Vi for the common-base configuration.
316
Chapter 7
BJT Transistor Modeling
re
The fact that the polarity of the voltage Vo as determined by the current Ic is the
same as defined by Fig. 7.19 (i.e., the negative side is at ground potential) reveals that
Vo and Vi are in phase for the common-base configuration. For an npn transistor in
the common-base configuration, the equivalence would appear as shown in Fig. 7.20.
Figure 7.20 Approximate model for a common-base npn transistor configuration.
For a common-base configuration of Fig. 7.17 with IE 4 mA, 0.98, and an ac
signal of 2 mV applied between the base and emitter terminals:
(a) Determine the input impedance.
(b) Calculate the voltage gain if a load of 0.56 k is connected to the output terminals.
(c) Find the output impedance and current gain.
EXAMPLE 7.4
Solution
26 mV
26 mV
(a) re 6.5 IE
4 mA
V
2 mV
(b) Ii Ie i 307.69 A
Zi
6.5 Vo Ic RL IeRL (0.98)(307.69 A)(0.56 k)
168.86 mV
Vo
168.86 mV
84.43
and Av Vi
2 mV
or from Eq. (7.14),
RL
(0.98)(0.56 k)
84.43
Av re
6.5 (c) Zo I
Ai o 0.98
Ii
as defined by Eq. (7.15)
Common Emitter Configuration
For the common-emitter configuration of Fig. 7.21a, the input terminals are the base
and emitter terminals, but the output set is now the collector and emitter terminals.
In addition, the emitter terminal is now common between the input and output ports
of the amplifier. Substituting the re equivalent circuit for the npn transistor will result
in the configuration of Fig. 7.21b. Note that the controlled-current source is still connected between the collector and base terminals and the diode between the base and
7.5 The re Transistor Model
317
re
Figure 7.21 (a) Common-emitter BJT transistor; (b) approximate model for the
configuration of Fig. 7.21a.
emitter terminals. In this configuration, the base current is the input current while the
output current is still Ic. Recall from Chapter 3 that the base and collector currents
are related by the following equation:
Ic Ib
(7.16)
The current through the diode is therefore determined by
Ie Ic Ib Ib Ib
Ie ( 1)Ib
and
(7.17)
However, since the ac beta is typically much greater than 1, we will use the following approximation for the current analysis:
Ie Ib
(7.18)
The input impedance is determined by the following ratio:
V
V
Zi i be
Ii
Ib
The voltage Vbe is across the diode resistance as shown in Fig. 7.22. The level of re
is still determined by the dc current IE. Using Ohm’s law gives
Vi Vbe Iere Ibre
Figure 7.22 Determining Zi using the approximate model.
318
Chapter 7
BJT Transistor Modeling
re
Substituting yields
V
Ir
Zi be be
Ib
Ib
Zi re
and
(7.19)
CE
In essence, Eq. (7.19) states that the input impedance for a situation such as shown
in Fig. 7.23 is beta times the value of re. In other words, a resistive element in the
emitter leg is reflected into the input circuit by a multiplying factor . For instance,
if re 6.5 as in Example 7.4 and 160 (quite typical), the input impedance has
increased to a level of
Zi re (160)(6.5 ) 1.04 k
For the common-emitter configuration, typical values of Zi defined by re
range from a few hundred ohms to the kilohm range, with maximums of
about 6–7 k.
Figure 7.23 Impact of re on
input impedance.
For the output impedance, the characteristics of interest are the output set of Fig.
7.24. Note that the slope of the curves increases with increase in collector current.
The steeper the slope, the less the level of output impedance (Zo). The re model of
Fig. 7.21 does not include an output impedance, but if available from a graphical
analysis or from data sheets, it can be included as shown in Fig. 7.25.
IC (mA)
Slope =
10
50 µA
1
ro
1
40 µA
8
30 µA
ro2 > ro1
20 µA
6
10 µA
4
I B = 0 µA
2
Slope =
0
Figure 7.25 Including ro in the
transistor equivalent circuit.
10
1
ro
2
20
VCE
Figure 7.24 Defining ro for the
common-emitter configuration.
For the common-emitter configuration, typical values of Zo are in the range of
40 to 50 k.
For the model of Fig. 7.25, if the applied signal is set to zero, the current Ic is
0 A and the output impedance is
Zo ro
(7.20)
CE
Of course, if the contribution due to ro is ignored as in the re model, the output impedance is defined by Zo .
The voltage gain for the common-emitter configuration will now be determined
for the configuration of Fig. 7.26 using the assumption that Zo . The effect of
including ro will be considered in Chapter 8. For the defined direction of Io and polarity of Vo,
Vo IoRL
7.5 The re Transistor Model
319
re
Figure 7.26 Determining the
voltage and current gain for the
common-emitter transistor amplifier.
The minus sign simply reflects the fact that the direction of Io in Fig. 7.26 would establish a voltage Vo with the opposite polarity. Continuing gives
Vo IoRL IcRL IbRL
Vi IiZi Ib re
and
so that
and
Vo
I RL
Av b
Vi
Ib re
RL
Av re
(7.21)
CE,ro The resulting minus sign for the voltage gain reveals that the output and input voltages are 180° out of phase.
The current gain for the configuration of Fig. 7.26:
I
Ib
I
Ai o c Ib
Ib
Ii
and
Ai CE,ro (7.22)
Using the facts that the input impedance is re, the collector current is Ib, and the
output impedance is ro, the equivalent model of Fig. 7.27 can be an effective tool in
the analysis to follow. For typical parameter values, the common-emitter configuration can be considered one that has a moderate level of input impedance, a high voltage and current gain, and an output impedance that may have to be included in the
network analysis.
Figure 7.27 re model for the
common-emitter transistor configuration.
EXAMPLE 7.5
320
Given 120 and IE 3.2 mA for a common-emitter configuration with ro ,
determine:
(a) Zi.
(b) Av if a load of 2 k is applied.
(c) Ai with the 2 k load.
Chapter 7
BJT Transistor Modeling
re
Solution
26 mV
26 mV
(a) re 8.125 IE
3.2 mA
and Zi re (120)(8.125 ) 975 RL
2 k
(b) Eq. (7.21): Av 246.15
re
8.125 I
(c) Ai o Ii
120
Common Collector Configuration
For the common-collector configuration, the model defined for the common-emitter
configuration of Fig. 7.21 is normally applied rather than defining a model for the
common-collector configuration. In subsequent chapters, a number of commoncollector configurations will be investigated and the impact of using the same model
will become quite apparent.
7.6 THE HYBRID EQUIVALENT MODEL
It was pointed out in Section 7.5 that the re model for a transistor is sensitive to the
dc level of operation of the amplifier. The result is an input resistance that will vary
with the dc operating point. For the hybrid equivalent model to be described in this
section, the parameters are defined at an operating point that may or may not reflect
the actual operating conditions of the amplifier. This is due to the fact that specification sheets cannot provide parameters for an equivalent circuit at every possible operating point. They must choose operating conditions that they believe reflect the general characteristics of the device. The hybrid parameters as shown in Fig. 7.28 are
drawn from the specification sheet for the 2N4400 transistor described in Chapter 3.
The values are provided at a dc collector current of 1 mA and a collector-to-emitter
voltage of 10 V. In addition, a range of values is provided for each parameter for guidance in the initial design or analysis of a system. One obvious advantage of the specification sheet listing is the immediate knowledge of typical levels for the parameters
of the device as compared to other transistors.
The quantities hie, hre, hfe, and hoe of Fig. 7.28 are called the hybrid parameters
and are the components of a small-signal equivalent circuit to be described shortly.
For years, the hybrid model with all its parameters was the chosen model for the educational and industrial communities. Presently, however, the re model is applied more
frequently, but often with the hoe parameter of the hybrid equivalent model to provide
Min.
Max.
hie
0.5
7.5
k
Voltage feedback ratio
(IC = 1 mA dc, VCE = 10 V dc, f = 1 kHz)
hre
0.1
8.0
104
Small-signal current gain
(IC = 1 mA dc, VCE = 10 V dc, f = 1 kHz) 2N4400
hfe
20
250
—
Output admittance
(IC = 1 mA dc, VCE = 10 V dc, f = 1 kHz)
hoe
1.0
30
1 S
Input impedance
(IC = 1 mA dc, VCE = 10 V dc, f = 1 kHz) 2N4400
Figure 7.28 Hybrid parameters for the 2N4400 transistor.
7.6 The Hybrid Equivalent Model
321
re
some measure for the output impedance. Since specification sheets do provide the hybrid parameters and the hybrid model continues to receive a good measure of attention, it is quite important that the hybrid model be covered in some detail in this book.
Once developed, the similarities between the re and hybrid models will be quite apparent. In fact, once the components of one are defined for a particular operating point,
the parameters of the other model are immediately available.
Our description of the hybrid equivalent model will begin with the general twoport system of Fig. 7.29. The following set of equations (7.23) is only one of a number of ways in which the four variables of Fig. 7.29 can be related. It is the most frequently employed in transistor circuit analysis, however, and therefore is discussed in
detail in this chapter.
Figure 7.29 Two-port system.
Vi h11Ii h12Vo
(7.23a)
Io h21Ii h22Vo
(7.23b)
The parameters relating the four variables are called h-parameters from the word
“hybrid.” The term hybrid was chosen because the mixture of variables (V and I ) in
each equation results in a “hybrid” set of units of measurement for the h-parameters.
A clearer understanding of what the various h-parameters represent and how we can
determine their magnitude can be developed by isolating each and examining the resulting relationship.
If we arbitrarily set Vo 0 (short circuit the output terminals) and solve for h11
in Eq. (7.23a), the following will result:
V
h11 i
Ii
Vo 0
ohms
(7.24)
The ratio indicates that the parameter h11 is an impedance parameter with the units
of ohms. Since it is the ratio of the input voltage to the input current with the output
terminals shorted, it is called the short-circuit input-impedance parameter. The subscript 11 of h11 defines the fact that the parameter is determined by a ratio of quantities measured at the input terminals.
If Ii is set equal to zero by opening the input leads, the following will result for
h12:
V
h12 i
Vo
Ii 0
unitless
(7.25)
The parameter h12, therefore, is the ratio of the input voltage to the output voltage
with the input current equal to zero. It has no units since it is a ratio of voltage levels and is called the open-circuit reverse transfer voltage ratio parameter. The subscript 12 of h12 reveals that the parameter is a transfer quantity determined by a ratio of input to output measurements. The first integer of the subscript defines the
322
Chapter 7
BJT Transistor Modeling
re
measured quantity to appear in the numerator; the second integer defines the source
of the quantity to appear in the denominator. The term reverse is included because
the ratio is an input voltage over an output voltage rather than the reverse ratio typically of interest.
If in Eq. (7.23b) Vo is equal to zero by again shorting the output terminals, the
following will result for h21:
I
h21 o
Ii
Vo0
unitless
(7.26)
Note that we now have the ratio of an output quantity to an input quantity. The term
forward will now be used rather than reverse as indicated for h12. The parameter h21
is the ratio of the output current to the input current with the output terminals shorted.
This parameter, like h12, has no units since it is the ratio of current levels. It is formally called the short-circuit forward transfer current ratio parameter. The subscript
21 again indicates that it is a transfer parameter with the output quantity in the numerator and the input quantity in the denominator.
The last parameter, h22, can be found by again opening the input leads to set
I1 0 and solving for h22 in Eq. (7.23b):
I
h22 o
Vo
Ii0
siemens
(7.27)
Since it is the ratio of the output current to the output voltage, it is the output conductance parameter and is measured in siemens (S). It is called the open-circuit output admittance parameter. The subscript 22 reveals that it is determined by a ratio of
output quantities.
Since each term of Eq. (7.23a) has the unit volt, let us apply Kirchhoff’s voltage
law “in reverse” to find a circuit that “fits” the equation. Performing this operation
will result in the circuit of Fig. 7.30. Since the parameter h11 has the unit ohm, it is
represented by a resistor in Fig. 7.30. The quantity h12 is dimensionless and therefore
simply appears as a multiplying factor of the “feedback” term in the input circuit.
Since each term of Eq. (7.23b) has the units of current, let us now apply Kirchhoff’s current law “in reverse” to obtain the circuit of Fig. 7.31. Since h22 has the
units of admittance, which for the transistor model is conductance, it is represented
by the resistor symbol. Keep in mind, however, that the resistance in ohms of this resistor is equal to the reciprocal of conductance (1/h22).
The complete “ac” equivalent circuit for the basic three-terminal linear device is
indicated in Fig. 7.32 with a new set of subscripts for the h-parameters. The notation
of Fig. 7.32 is of a more practical nature since it relates the h-parameters to the resulting ratio obtained in the last few paragraphs. The choice of letters is obvious from
the following listing:
Figure 7.30 Hybrid input
equivalent circuit.
h11 → input resistance → hi
h12 → reverse transfer voltage ratio → hr
Figure 7.31 Hybrid output
equivalent circuit.
Figure 7.32 Complete hybrid
equivalent circuit.
7.6 The Hybrid Equivalent Model
323
re
h21 → forward transfer current ratio → hf
h22 → output conductance → ho
The circuit of Fig. 7.32 is applicable to any linear three-terminal electronic device or
system with no internal independent sources. For the transistor, therefore, even though
it has three basic configurations, they are all three-terminal configurations, so that the
resulting equivalent circuit will have the same format as shown in Fig. 7.32. In each
case, the bottom of the input and output sections of the network of Fig. 7.32 can be
connected as shown in Fig. 7.33 since the potential level is the same. Essentially,
therefore, the transistor model is a three-terminal two-port system. The h-parameters,
however, will change with each configuration. To distinguish which parameter has
been used or which is available, a second subscript has been added to the h-parameter notation. For the common-base configuration, the lowercase letter b was added,
while for the common-emitter and common-collector configurations, the letters e and
c were added, respectively. The hybrid equivalent network for the common-emitter
configuration appears with the standard notation in Fig. 7.33. Note that Ii Ib, Io Ic, and through an application of Kirchhoff’s current law, Ie Ib Ic. The input voltage is now Vbe, with the output voltage Vce. For the common-base configuration of
Fig. 7.34, Ii Ie, Io Ic with Veb Vi and Vcb Vo. The networks of Figs. 7.33 and
7.34 are applicable for pnp or npn transistors.
The fact that both a Thévenin and Norton circuit appear in the circuit of Fig. 7.32
was further impetus for calling the resultant circuit a hybrid equivalent circuit. Two
additional transistor equivalent circuits, not to be discussed in this text, called the
Figure 7.33 Common-emitter configuration: (a) graphical symbol; (b) hybrid
equivalent circuit.
Figure 7.34 Common-base configuration: (a) graphical symbol; (b) hybrid
equivalent circuit.
324
Chapter 7
BJT Transistor Modeling
re
z-parameter and y-parameter equivalent circuits, use either the voltage source or the
current source, but not both, in the same equivalent circuit. In Section 7.7, the magnitudes of the various parameters will be found from the transistor characteristics in
the region of operation resulting in the desired small-signal equivalent network for
the transistor.
For the common-emitter and common-base configurations, the magnitude of hr
and ho is often such that the results obtained for the important parameters such as Zi,
Zo, Av, and Ai are only slightly affected if they (hr and ho) are not included in the
model.
Since hr is normally a relatively small quantity, its removal is approximated by
hr 0 and hrVo 0, resulting in a short-circuit equivalent for the feedback element
as shown in Fig. 7.35. The resistance determined by 1/ho is often large enough to be
ignored in comparison to a parallel load, permitting its replacement by an opencircuit equivalent for the CE and CB models, as shown in Fig. 7.35.
The resulting equivalent of Fig. 7.36 is quite similar to the general structure of
the common-base and common-emitter equivalent circuits obtained with the re model.
In fact, the hybrid equivalent and the re models for each configuration have been repeated in Fig. 7.37 for comparison. It should be reasonably clear from Fig. 7.37a that
Ii
Io
+
+
hi
Vi
Vo
h f Ii
–
–
Figure 7.36 Approximate hybrid equivalent model.
Figure 7.35 Effect of removing hre and hoe from the
hybrid equivalent circuit.
Ib
Ic
Ib
c
b
h ie
Ic
c
b
β re
h fe Ib
e
e
β Ib
e
e
(a)
Ic
Ie
e
c
hib
b
Ic
Ie
e
c
re
h f b Ib
b
b
α Ie
e
(b)
Figure 7.37 Hybrid versus re model: (a) common-emitter configuration;
(b) common-base configuration.
7.6 The Hybrid Equivalent Model
325
re
and
hie re
(7.28)
hfe (7.29)
ac
From Fig. 7.37b,
and
hib re
(7.30)
hfb 1
(7.31)
In particular, note that the minus sign in Eq. (7.31) accounts for the fact that the current source of the standard hybrid equivalent circuit is pointing down rather than in
the actual direction as shown in the re model of Fig. 7.37b.
EXAMPLE 7.6
Given IE 2.5 mA, hfe 140, hoe 20 S ( mho), and hob 0.5
(a) The common-emitter hybrid equivalent circuit.
(b) The common-base re model.
S, determine:
Solution
26 mV
26 mV
(a) re 10.4 2.5 mA
IE
hie re (140)(10.4 ) 1.456 k
1
1
ro 50 k
hoe
20 S
Note Fig. 7.38.
c
b
Ib
h ie
Figure 7.38 Common-emitter
hybrid equivalent circuit for the
parameters of example 7.6.
1.456 kΩ
140 Ib
e
e
(b) re 10.4 1
1
ro 2 M
hob
0.5 S
1,
Note Fig. 7.39.
Figure 7.39 Common-base re
model for the parameters of example 7.6.
326
1
= 50 kΩ
h oe
Chapter 7
BJT Transistor Modeling
re
A series of equations relating the parameters of each configuration for the hybrid
equivalent circuit is provided in Appendix A. In Section 7.8, we demonstrate that the
hybrid parameter hfe ( ac) is the least sensitive of the hybrid parameters to a change
in collector current. Assuming, therefore, that hfe is a constant for the range of
interest, is a fairly good approximation. It is hie re that will vary significantly with
IC and should be determined at operating levels, since it can have a real impact on
the gain levels of a transistor amplifier.
7.7 GRAPHICAL DETERMINATION OF
THE h-PARAMETERS
Using partial derivatives (calculus), it can be shown that the magnitude of the hparameters for the small-signal transistor equivalent circuit in the region of operation
for the common-emitter configuration can be found using the following equations:*
v
vbe
vbe
hie i ii
ib
ib
v
vbe
vbe
hre i vo
vce
vce
i
i
i
hfe o c c
ib
ib
ii
VCE constant
IB constant
VCE constant
io
ic
ic
hoe vo
vce
vce
IB constant
(ohms)
(7.32)
(unitless)
(7.33)
(unitless)
(siemens)
(7.34)
(7.35)
In each case, the symbol refers to a small change in that quantity around the
quiescent point of operation. In other words, the h-parameters are determined in the
region of operation for the applied signal so that the equivalent circuit will be the
most accurate available. The constant values of VCE and IB in each case refer to a condition that must be met when the various parameters are determined from the characteristics of the transistor. For the common-base and common-collector configurations, the proper equation can be obtained by simply substituting the proper values
of vi, vo, ii, and io.
The parameters hie and hre are determined from the input or base characteristics,
while the parameters hfe and hoe are obtained from the output or collector characteristics. Since hfe is usually the parameter of greatest interest, we shall discuss the operations involved with equations, such as Eqs. (7.32) through (7.35), for this parameter first. The first step in determining any of the four hybrid parameters is to find
the quiescent point of operation as indicated in Fig. 7.40. In Eq. (7.34) the condition
VCE constant requires that the changes in base current and collector current be taken
along a vertical straight line drawn through the Q-point representing a fixed collector-to-emitter voltage. Equation (7.34) then requires that a small change in collector
current be divided by the corresponding change in base current. For the greatest accuracy, these changes should be made as small as possible.
*The partial derivative vi/ii provides a measure of the instantaneous change in vi due to an instantaneous
change in ii.
7.7 Graphical Determination of the h-Parameters
327
re
Figure 7.40 hfe determination.
In Fig. 7.40, the change in ib was chosen to extend from IB1 to IB2 along the perpendicular straight line at VCE. The corresponding change in ic is then found by drawing the horizontal lines from the intersections of IB1 and IB2 with VCE constant to
the vertical axis. All that remains is to substitute the resultant changes of ib and ic into
Eq. (7.34). That is,
i
hfe c
ib
(2.7 1.7) mA
(20 10) A
VCE constant
VCE 8.4 V
103
100
10 106
In Fig. 7.41, a straight line is drawn tangent to the curve IB through the Q-point
to establish a line IB constant as required by Eq. (7.35) for hoe. A change in vCE
was then chosen and the corresponding change in iC determined by drawing the horizontal lines to the vertical axis at the intersections on the IB constant line. Substituting into Eq. (7.35), we get
Figure 7.41 hoe determination.
328
Chapter 7
BJT Transistor Modeling
re
(2.2 2.1) mA
ic
hoe (10 7) V IB 15
vce IB constant
A
3
0.1 10
33 A/V 33 106 S 33 S
3
To determine the parameters hie and hre the Q-point must first be found on the input or base characteristics as indicated in Fig. 7.42. For hie, a line is drawn tangent
to the curve VCE 8.4 V through the Q-point to establish a line VCE constant as
required by Eq. (7.32). A small change in vbe was then chosen, resulting in a corresponding change in ib. Substituting into Eq. (7.32), we get
vbe
hie ib
VCE constant
(733 718) mV
(20 10) A
VCE 8.4 V
15 103
1.5 k
10 106
Figure 7.42 hie determination.
The last parameter, hre, can be found by first drawing a horizontal line through
the Q-point at IB 15 A. The natural choice then is to pick a change in vCE and
find the resulting change in vBE as shown in Fig. 7.43.
Substituting into Eq. (7.33), we get
vbe
(733 725) mV
8 103
4 104
hre vce IB constant
(20 0) V
20
For the transistor whose characteristics have appeared in Figs. 7.40 through 7.43,
the resulting hybrid small-signal equivalent circuit is shown in Fig. 7.44.
7.7 Graphical Determination of the h-Parameters
329
re
Figure 7.43 hre determination.
Figure 7.44 Complete hybrid
equivalent circuit for a transistor
having the characteristics that
appear in Figs. 7.40 through 7.43.
As mentioned earlier, the hybrid parameters for the common-base and commoncollector configurations can be found using the same basic equations with the proper
variables and characteristics.
Table 7.1 lists typical parameter values in each of the three configurations for the
broad range of transistors available today. The minus sign indicates that in Eq. (7.34)
as one quantity increased in magnitude, within the change chosen, the other decreased
in magnitude.
TABLE 7.1 Typical Parameter Values for the CE, CC, and CB
Transistor Configurations
Parameter
hi
hr
hf
ho
1/ho
330
Chapter 7
CE
CC
CB
01 k
02.5 104
50
25 A/V
40 k
01 k
1
50
25 A/V
40 k
20 03.0 104
0.98
00.5 A/V0
2 M
BJT Transistor Modeling
re
Note in retrospect (Section 3.5: Transistor Amplifying Action) that the input resistance of the common-base configuration is low, while the output resistance is high.
Consider also that the short-circuit current gain is very close to 1. For the commonemitter and common-collector configurations, note that the input resistance is much
higher than that of the common-base configuration and that the ratio of output to input resistance is about 401. Consider also for the common-emitter and common-base
configurations that hr is very small in magnitude. Transistors are available today with
values of hfe that vary from 20 to 600. For any transistor, the region of operation and
conditions under which it is being used will have an effect on the various h-parameters. The effect of temperature and collector current and voltage on the h-parameters
is discussed in Section 7.8.
7.8 VARIATIONS OF TRANSISTOR
PARAMETERS
There are a large number of curves that can be drawn to show the variations of the
h-parameters with temperature, frequency, voltage, and current. The most interesting
and useful at this stage of the development include the h-parameter variations with
junction temperature and collector voltage and current.
In Fig. 7.45, the effect of the collector current on the h-parameter has been indicated. Take careful note of the logarithmic scale on the vertical and horizontal axes.
Logarithmic scales will be examined in Chapter 11. The parameters have all been normalized to unity so that the relative change in magnitude with collector current can
easily be determined. On every set of curves, such as in Fig. 7.46, the operating point
at which the parameters were found is always indicated. For this particular situation,
the quiescent point is at the intersection of VCE 5.0 V and IC 1.0 mA. Since the
frequency and temperature of operation will also affect the h-parameters, these quantities are also indicated on the curves. At 0.1 mA, hfe is about 0.5 or 50% of its value
at 1.0 mA, while at 3 mA, it is 1.5 or 150% of that value. In other words, if hfe 50
at IC 1.0 mA, hfe has changed from a value of 0.5(50) 25 to 1.5(50) 75, with
Figure 7.45 Hybrid parameter variations with collector current.
7.8 Variations of Transistor Parameters
331
re
Figure 7.46 Hybrid parameter variations with collector–emitter potential.
a change of IC from 0.1 to 3 mA. Consider, however, the point of operation at
IC 50 mA. The magnitude of hre is now approximately 11 times that at the defined
Q-point, a magnitude that may not permit eliminating this parameter from the equivalent circuit. The parameter hoe is approximately 35 times the normalized value. This
increase in hoe will decrease the magnitude of the output resistance of the transistor
to a point where it may approach the magnitude of the load resistor. There would then
be no justification in eliminating hoe from the equivalent circuit on an approximate
basis.
In Fig. 7.46, the variation in magnitude of the h-parameters on a normalized basis has been indicated with changes in collector voltage. This set of curves was normalized at the same operating point of the transistor discussed in Fig. 7.45 so that a
comparison between the two sets of curves can be made. Note that hie and hfe are relatively steady in magnitude while hoe and hre are much larger to the left and right of
the chosen operating point. In other words, hoe and hre are much more sensitive to
changes in collector voltage than are hie and hfe.
It is interesting to note from Figs. 7.45 and 7.46 that the value of hfe appears to
change the least. Therefore, the specific value of current gain, whether hfe or , can,
on an approximate and relative basis, be considered fairly constant for the range of
collector current and voltage.
The value of hie re does vary considerably with collector current as one might
expect due to the sensitivity of re to emitter (IE IC) current. It is therefore a quantity that should be determined as close to operating conditions as possible. For values below the specified VCE, hre is fairly constant, but it does increase measurably for
higher values. It is indeed fortunate that for most applications the magnitude of hre
and hoe are such that they can usually be ignored. They are quite sensitive to collector current and collector-to-emitter voltage.
In Fig. 7.47, the variation in h-parameters has been plotted for changes in junction temperature. The normalization value is taken to be room temperature: T 25°C.
The horizontal scale is a linear scale rather than a logarithmic scale as was employed
for Figs. 7.45 and 7.46. In general, all the parameters increase in magnitude with temperature. The parameter least affected, however, is hoe, while the input impedance hie
changes at the greatest rate. The fact that hfe will change from 50% of its normalized
value at 50°C to 150% of its normalized value at 150°C indicates clearly that the
operating temperature must be carefully considered in the design of transistor circuits.
332
Chapter 7
BJT Transistor Modeling
re
Figure 7.47 Hybrid parameter
variations with temperature.
PROBLEMS
§ 7.2 Amplification in the AC Domain
1. (a) What is the expected amplification of a BJT transistor amplifier if the dc supply is set to
zero volts?
(b) What will happen to the output ac signal if the dc level is insufficient? Sketch the effect on
the waveform.
(c) What is the conversion efficiency of an amplifier in which the effective value of the current through a 2.2-k load is 5 mA and the drain on the 18-V dc supply is 3.8 mA?
2. Can you think of an analogy that would explain the importance of the dc level on the resulting
ac gain?
§
7.3 BJT Transistor Modeling
3. What is the reactance of a 10- F capacitor at a frequency of 1 kHz? For networks in which the
resistor levels are typically in the kilohm range, is it a good assumption to use the short-circuit
equivalence for the conditions just described? How about at 100 kHz?
4. Given the common-base configuration of Fig. 7.48, sketch the ac equivalent using the notation
for the transistor model appearing in Fig. 7.5.
Figure 7.48 Problem 4
Problems
333
re
5. (a) Describe the differences between the re and hybrid equivalent models for a BJT transistor.
(b) For each model, list the conditions under which it should be applied.
§
7.4 The Important Parameters: Zi, Zo, Av, Ai
6. (a) For the configuration of Fig. 7.7, determine Zi if Vs 40 mV, Rsense 0.5 k, and Ii 20 A.
(b) Using the results of part (a), determine Vi if the applied source is changed to 12 mV with
an internal resistance of 0.4 k.
7. (a) For the network of Fig. 7.10, determine Zo if V 600 mV, Rsense 10 k, and Io 10 A.
(b) Using the Zo obtained in part (a), determine IL for the configuration of Fig. 7.11 if RL 2.2 k and Iamplifier 6 mA.
8. Given the BJT configuration of Fig. 7.49, determine:
(a)
(b)
(c)
(d)
Vi.
Zi.
AvNL.
Avs.
Ii = 10 µA
+
Vs
Zi
18 mV
+
+
0.6 kΩ
Vi
–
–
BJT
transistor
amplifier
Aυ
Vo = 3.6 V
–
NL
Figure 7.49 Problem 8
9. For the BJT amplifier of Fig. 7.50, determine:
(a)
(b)
(c)
(d)
(e)
(f)
Ii.
Zi.
Vo.
Io.
Ai using the results of parts (a) and (d).
Ai using Eq. (7.10).
Ii
1 kΩ
+
Vs
Zi
12 mV
–
Io
Vi = 4 mV
–
Figure 7.50 Problem 9
334
Chapter 7
+
+
BJT Transistor Modeling
BJT
transistor
amplifier
Aυ = –180
Vo
–
RL
0.51 kΩ
re
§
7.5 The re Transistor Model
10. For the common-base configuration of Fig. 7.17, an ac signal of 10 mV is applied, resulting in
an emitter current of 0.5 mA. If 0.980, determine:
(a) Zi.
(b) Vo if RL 1.2 k.
(c) Av Vo / Vi.
(d) Zo with ro .
(e) Ai Io /Ii.
(f) Ib.
11. For the common-base configuration of Fig. 7.17, the emitter current is 3.2 mA and
Determine the following if the applied voltage is 48 mV and the load is 2.2 k.
(a) re.
(b) Zi.
(c) Ic.
(d) Vo.
(e) Av.
(f) Ib.
is 0.99.
12. Using the model of Fig. 7.27, determine the following for a common-emitter amplifier if
80, IE (dc) 2 mA, and ro 40 k.
(a) Zi.
(b) Ib.
(c) Ai Io / Ii IL /Ib if RL 1.2 k.
(d) Av if RL 1.2 k.
13. The input impedance to a common-emitter transistor amplifier is 1.2 k with
ro 50 k, and RL 2.7 k. Determine:
(a) re.
(b) Ib if Vi 30 mV.
(c) Ic.
(d) Ai Io/Ii IL/Ib.
(e) Av Vo/Vi.
§
140,
7.6 The Hybrid Equivalent Model
14. Given IE (dc) 1.2 mA, 120, and ro 40 k, sketch the:
(a) Common-emitter hybrid equivalent model.
(b) Common-emitter re equivalent model.
(c) Common-base hybrid equivalent model.
(d) Common-base re equivalent model.
15. Given hie 2.4 k, hfe 100, hre 4104, and hoe 25 S, sketch the:
(a) Common-emitter hybrid equivalent model.
(b) Common-emitter re equivalent model.
(c) Common-base hybrid equivalent model.
(d) Common-base re equivalent model.
16. Redraw the common-emitter network of Fig. 7.3 for the ac response with the approximate hybrid equivalent model substituted between the appropriate terminals.
Problems
335
re
17. Redraw the network of Fig. 7.51 for the ac response with the re model inserted between the appropriate terminals. Include ro.
18. Redraw the network of Fig. 7.52 for the ac response with the re model inserted between the appropriate terminals. Include ro.
Figure 7.52 Problem 18
Figure 7.51 Problem 17
19. Given the typical values of hie 1 k, hre 2 104, and Av 160 for the input configuration of Fig. 7.53:
(a) Determine Vo in terms of Vi.
(b) Calculate Ib in terms of Vi.
(c) Calculate Ib if hreVo is ignored.
(d) Determine the percent difference in Ib using the following equation:
Ib(without hre) Ib(with hre)
% difference in Ib 100%
Ib(without hre)
Figure 7.53 Problems 19, 21
(e) Is it a valid approach to ignore the effects of hreVo for the typical values employed in this
example?
20. Given the typical values of RL 2.2 k and hoe 20 S, is it a good approximation to ignore
the effects of 1/hoe on the total load impedance? What is the percent difference in total loading
on the transistor using the following equation?
RL RL(1/hoe)
100%
% difference in total load RL
21. Repeat Problem 19 using the average values of the parameters of Fig. 7.28 with Av 180.
22. Repeat Problem 20 for RL 3.3 k and the average value of hoe in Fig. 7.28.
§
7.7 Graphical Determination of the h-Parameters
23. (a) Using the characteristics of Fig. 7.40, determine hfe at IC 6 mA and VCE 5 V.
(b) Repeat part (a) at IC 1 mA and VCE 15 V.
336
Chapter 7
BJT Transistor Modeling
re
24. (a) Using the characteristics of Fig. 7.41, determine hoe at IC 6 mA and VCE 5 V.
(b) Repeat part (a) at IC 1 mA and VCE 15 V.
25. (a) Using the characteristics of Fig. 7.42, determine hie at IB 20 A and VCE 20 V.
(b) Repeat part (a) at IB 5 A and VCE 10 V.
26. (a) Using the characteristics of Fig. 7.43, determine hre at IB 20 A.
(b) Repeat part (a) at IB 30 A.
* 27. Using the characteristics of Figs. 7.40 and 7.42, determine the approximate CE hybrid equivalent model at IB 25 A and VCE 12.5 V.
* 28. Determine the CE re model at IB 25 A and VCE 12.5 V using the characteristics of Figs.
7.40 and 7.42.
* 29. Using the results of Fig. 7.44, sketch the re equivalent model for the transistor having the characteristics appearing in Figs. 7.40 through 7.43. Include ro.
§
7.8 Variations of Transistor Parameters
For Problems 30 through 34, use Figs. 7.45 through 7.47.
30. (a) Using Fig. 7.45, determine the magnitude of the percent change in hfe for an IC change from
0.2 mA to 1 mA using the equation
hfe(0.2 mA) hfe(1 mA)
% change 100%
hfe(0.2 mA)
(b) Repeat part (a) for an IC change from 1 mA to 5 mA.
31. Repeat Problem 30 for hie (same changes in IC).
32. (a) If hoe 20 S at IC 1 mA on Fig. 7.45, what is the approximate value of hoe at IC 0.2 mA?
(b) Determine its resistive value at 0.2 mA and compare to a resistive load of 6.8 k. Is it a
good approximation to ignore the effects of 1/hoe in this case?
33. (a) If hoe 20 S at IC 1 mA on Fig. 7.45, what is the approximate value of hoe at IC 10 mA?
(b) Determine its resistive value at 10 mA and compare to a resistive load of 6.8 k. Is it a
good approximation to ignore the effects of 1/hoe in this case?
34. (a) If hre 2 104 at IC 1 mA on Fig. 7.45, determine the approximate value of hre at
0.1 mA.
(b) Using the value of hre determined in part (a), can hre be ignored as a good approximation
if Av 210?
* 35. (a) Reviewing the characteristics of Fig. 7.45, which parameter changed the least for the full
range of collector current?
(b) Which parameter changed the most?
(c) What are the maximum and minimum values of 1/hoe? Is the approximation 1/hoeRL RL
more appropriate at high or low levels of collector current?
(d) In which region of current spectrum is the approximation hreVce 0 the most appropriate?
36. (a) Reviewing the characteristics of Fig. 7.47, which parameter changed the most with increase
in temperature?
(b) Which changed the least?
(c) What are the maximum and minimum values of hfe? Is the change in magnitude significant? Was it expected?
(d) How does re vary with increase in temperature? Simply calculate its level at three or four
points and compare their magnitudes.
(e) In which temperature range do the parameters change the least?
*Please Note: Asterisks indicate more difficult problems.
Problems
337
CHAPTER
8
BJT Small-Signal
Analysis
8.1 INTRODUCTION
The transistor models introduced in Chapter 7 will now be used to perform a smallsignal ac analysis of a number of standard transistor network configurations. The networks analyzed represent the majority of those appearing in practice today. Modifications of the standard configurations will be relatively easy to examine once the
content of this chapter is reviewed and understood.
Since the re model is sensitive to the actual point of operation, it will be our primary model for the analysis to be performed. For each configuration, however, the
effect of an output impedance is examined as provided by the hoe parameter of the
hybrid equivalent model. To demonstrate the similarities in analysis that exist between
models, a section is devoted to the small-signal analysis of BJT networks using solely
the hybrid equivalent model. The analysis of this chapter does not include a load resistance RL or source resistance Rs. The effect of both parameters is reserved for a
systems approach in Chapter 10.
The computer analysis section includes a brief description of the transistor model
employed in the PSpice software package. It demonstrates the range and depth of the
computer analysis systems available today and how relatively easy it is to enter a complex network and print out the desired results.
8.2 COMMON-EMITTER FIXED-BIAS
CONFIGURATION
The first configuration to be analyzed in detail is the common-emitter fixed-bias network of Fig. 8.1. Note that the input signal Vi is applied to the base of the transistor
while the output Vo is off the collector. In addition, recognize that the input current
Ii is not the base current but the source current, while the output current Io is the collector current. The small-signal ac analysis begins by removing the dc effects of VCC
and replacing the dc blocking capacitors C1 and C2 by short-circuit equivalents, resulting in the network of Fig. 8.2.
Note in Fig. 8.2 that the common ground of the dc supply and the transistor emitter terminal permits the relocation of RB and RC in parallel with the input and output
sections of the transistor, respectively. In addition, note the placement of the important network parameters Zi, Zo, Ii, and Io on the redrawn network. Substituting the re
model for the common-emitter configuration of Fig. 8.2 will result in the network of
Fig. 8.3.
338
VCC
RC
RB
Io
C
C
Ii
Vo
C2
B
Vi
Io
B
Vi
RC
Zo
C1
Zi
Vo
Ii
RB
Zi
E
Figure 8.1 Common-emitter fixed-bias configuration.
Zo
E
Figure 8.2 Network of Figure 8.1 following
the removal of the effects of VCC, C1, and C 2.
Ib
Ii
+Z
Ic
c
b
Io
i
Vi
Figure 8.3 Substituting the re
model into the network of Fig.
8.2.
β re
RB
β Ib
ro
+
Vo
RC
–
–
Zo
The next step is to determine , re, and ro. The magnitude of is typically obtained from a specification sheet or by direct measurement using a curve tracer or
transistor testing instrument. The value of re must be determined from a dc analysis
of the system, and the magnitude of ro is typically obtained from the specification
sheet or characteristics. Assuming that , re, and ro have been determined will result
in the following equations for the important two-port characteristics of the system.
Zi: Figure 8.3 clearly reveals that
Zi RBre
ohms
(8.1)
For the majority of situations RB is greater than re by more than a factor of 10
(recall from the analysis of parallel elements that the total resistance of two parallel
resistors is always less than the smallest and very close to the smallest if one is much
larger than the other), permitting the following approximation:
Zi re
ohms
(8.2)
RB10re
Zo: Recall that the output impedance of any system is defined as the impedance
Zo determined when Vi 0. For Fig. 8.3, when Vi 0, Ii Ib 0, resulting in an
open-circuit equivalence for the current source. The result is the configuration of
Fig. 8.4.
Zo RCro
ohms
Zo
ro
RC
(8.3)
If ro 10 RD, the approximation RCro RC is frequently applied and
Zo RC
8.2
ro10RC
(8.4)
Common-Emitter Fixed-Bias Configuration
Figure 8.4 Determining Zo for
the network of Fig. 8.3.
339
Av:
The resistors ro and RC are in parallel,
and
Vo Ib(RC ro)
but
Vi
Ib rc
so that
Vi
Vo (RC ro)
re
and
Vo
(RC ro)
Av re
Vi
(8.5)
RC
Av re
(8.6)
If ro 10RC,
ro10RC
Note the explicit absence of in Eqs. (8.5 and 8.6), although we recognize that must be utilized to determine re.
Ai: The current gain is determined in the following manner: Applying the current-divider rule to the input and output circuits,
with
(ro)(Ib)
Io ro RC
and
Io
ro
ro RC
Ib
(RB)(Ii)
Ib RB re
or
RB
Ib
RB re
Ii
The result is
RB
Io
Io Ib
ro
Ai ro RC RB re
Ii
Ib Ii
and
RBro
Io
Ai (ro RC)(RB re)
Ii
(8.7)
which is certainly an unwieldy, complex expression.
However, if ro 10RC and RB 10re, which is often the case,
Io
RBro
Ai Ii
(ro)(RB)
and
Ai (8.8)
ro10RC, RB10re
The complexity of Eq. (8.7) suggests that we may want to return to an equation such
as Eq. (7.10), which utilizes Ao and Zi. That is,
Zi
Ai Av
RC
(8.9)
Phase Relationship: The negative sign in the resulting equation for Av reveals
that a 180° phase shift occurs between the input and output signals, as shown in Fig.
8.5.
340
Chapter 8
BJT Small-Signal Analysis
VCC
Vo
RC
RB
Vi
0
Vo
t
0
t
Vi
Figure 8.5 Demonstrating the
180° phase shift between input
and output waveforms.
For the network of Fig. 8.6:
(a) Determine re.
(b) Find Zi (with ro ).
(c) Calculate Zo (with ro ).
(d) Determine Av (with ro ).
(e) Find Ai (with ro ).
(f) Repeat parts (c) through (e) including ro 50 k in all calculations and compare
results.
EXAMPLE 8.1
12 V
3 kΩ
470 kΩ
Ii
Vi
10 µ F
Io
Vo
10 µ F
β = 100
ro = 50 kΩ
Zo
Zi
Figure 8.6
Example 8.1.
Solution
(a) DC analysis:
VCC VBE
12 V 0.7 V
IB 24.04 A
RB
470 k
IE ( 1)IB (101)(24.04 A) 2.428 mA
26 mV
26 mV
re 10.71 IE
2.428 mA
(b) re (100)(10.71 ) 1.071 k
Zi RBre 470 k1.071 k 1.069 k
(c) Zo RC 3 k
RC
3 k
(d) Av 280.11
re
10.71 (e) Since RB 10re(470 k 10.71 k)
Ai 100
8.2
Common-Emitter Fixed-Bias Configuration
341
(f) Zo roRC 50 k3 k 2.83 k vs. 3 k
2.83 k
roRC
Av 264.24 vs. 280.11
10.71 re
(100)(470 k)(50 k)
RBro
Ai (ro RC)(RB re)
(50 k 3 k)(470 k 1.071 k)
94.13 vs. 100
As a check:
(264.24)(1.069 k)
Zi
Ai Av 94.16
3 k
RC
which differs slightly only due to the accuracy carried through the calculations.
8.3 VOLTAGE-DIVIDER BIAS
The next configuration to be analyzed is the voltage-divider bias network of Fig. 8.7.
Recall that the name of the configuration is a result of the voltage-divider bias at the
input side to determine the dc level of VB.
VCC
Io
RC
R1
Vo
C
Ii
C2
B
Vi
Zo
C1
E
Zi
R2
RE
CE
Figure 8.7 Voltage-divider bias
configuration.
Substituting the re equivalent circuit will result in the network of Fig. 8.8. Note
the absence of RE due to the low-impedance shorting effect of the bypass capacitor,
CE. That is, at the frequency (or frequencies) of operation, the reactance of the capacitor is so small compared to RE that it is treated as a short circuit across RE. When
Ii
b
c
Ib
+
Vi
+
Io
Zi
R1
–
β re
R2
e
β Ib
ro
Vo
RC
e
Zo
–
R'
Figure 8.8 Substituting the re equivalent circuit into the ac equivalent network of Fig. 8.7.
342
Chapter 8
BJT Small-Signal Analysis
VCC is set to zero, it places one end of R1 and RC at ground potential as shown in Fig.
8.8. In addition, note that R1 and R2 remain part of the input circuit while RC is part
of the output circuit. The parallel combination of R1 and R2 is defined by
R1R2
R R1R2 R1 R2
(8.10)
Zi R re
(8.11)
Zi: From Fig. 8.8,
Zo: From Fig. 8.8 with Vi set to 0 V resulting in Ib 0
Zo RC ro
A and Ib 0 mA,
(8.12)
If ro 10RC,
Zo RC
Av :
(8.13)
ro10RC
Since RC and ro are in parallel,
Vo (Ib)(RC ro)
and
so that
and
Vi
Ib re
Vi
Vo (RC ro)
re
Vo
RC ro
Av Vi
re
(8.14)
which you will note is an exact duplicate of the equation obtained for the fixed-bias
configuration.
For ro 10RC,
Vo
RC
Av Vi
re
(8.15)
ro10RC
Ai: Since the network of Fig. 8.8 is so similar to that of Fig. 8.3 except for the
fact that R R1R2 RB, the equation for the current gain will have the same format as Eq. (8.7). That is,
R ro
Io
Ai (ro RC)(R re)
Ii
(8.16)
For ro 10RC,
Io
R ro
Ai Ii
ro(R re)
and
Io
R
Ai Ii
R re
(8.17)
ro10RC
8.3
Voltage-Divider Bias
343
And if R 10re,
Io
R
Ai Ii
R
Io
Ai Ii
and
(8.18)
ro10RC, R 10re
As an option,
Zi
Ai Av
RC
Phase relationship:
between Vo and Vi.
EXAMPLE 8.2
For
(a)
(b)
(c)
(d)
(e)
(f)
(8.19)
The negative sign of Eq. (8.14) reveals a 180° phase shift
the network of Fig. 8.9, determine:
re.
Zi.
Zo (ro ).
Av (ro ).
Ai (ro ).
The parameters of parts (b) through (e) if ro 1/hoe 50 k and compare results.
22 V
Io
6.8 kΩ
10 µF
56 kΩ
Vo
10 µ F
β = 90
Vi
Zo
Ii
8.2 kΩ
Zi
1.5 kΩ
20 µ F
Figure 8.9 Example 8.2.
Solution
(a) DC: Testing RE
10R2
(90)(1.5 k)
135 k
10(8.2 k)
82 k (satisfied)
Using the approximate approach,
R2
(8.2 k)(22 V)
VB VCC 2.81 V
R1 R2
56 k 8.2 k
VE VB VBE 2.81 V 0.7 V 2.11 V
344
Chapter 8
BJT Small-Signal Analysis
VE
2.11 V
IE 1.41 mA
RE
1.5 k
26 mV
26 mV
re 18.44 IE
1.41 mA
(b) R R1R2 (56 k)(8.2 k) 7.15 k
Zi R re 7.15 k(90)(18.44 ) 7.15 k1.66 k
1.35 k
(c) Zo RC 6.8 k
6.8 k
RC
(d) Av 368.76
18.44 re
(e) The condition R 10re (7.15 k 10(1.66 k) 16.6 k is not
satisfied. Therefore,
(90)(7.15 k)
R
Ai 7.15 k 1.66 k 73.04
R re
(f) Zi 1.35 k
Zo RCro 6.8 k50 k 5.98 k vs. 6.8 k
RC ro
5.98 k
Av 324.3 vs. 368.76
re
18.44 The condition
ro 10RC (50 k 10(6.8 k) 68 k)
is not satisfied. Therefore,
R ro
(90)(7.15 k)(50 k)
Ai (ro RC)(R re)
(50 k 6.8 k)(7.15 k 1.66 k)
64.3 vs. 73.04
There was a measurable difference in the results for Zo, Av, and Ai because the
condition ro 10RC was not satisfied.
8.4 CE EMITTER-BIAS CONFIGURATION
The networks examined in this section include an emitter resistor that may or may
not be bypassed in the ac domain. We will first consider the unbypassed situation and
then modify the resulting equations for the bypassed configuration.
Unbypassed
The most fundamental of unbypassed configurations appears in Fig. 8.10. The re
equivalent model is substituted in Fig. 8.11, but note the absence of the resistance ro.
The effect of ro is to make the analysis a great deal more complicated, and considering the fact that in most situations its effect can be ignored, it will not be included in
the current analysis. However, the effect of ro will be discussed later in this section.
Applying Kirchhoff’s voltage law to the input side of Fig. 8.11 will result in
Vi Ibre Ie RE
or
Vi Ibre ( 1)IbRE
8.4
CE Emitter-Bias Configuration
345
VCC
Ii
b
c
Ib
+
+
β re
RC
RB
Io
Io
Vo
Ii
β Ib
Zi
C2
Zo
Zb
Vi
RB
RC
Vo
e
Vi
Ie = ( β + 1)Ib
C1
RE
Zo
RE
–
Zi
Figure 8.10 CE emitter-bias configuration.
–
Figure 8.11 Substituting the re equivalent circuit into the ac equivalent network of Fig. 8.10.
and the input impedance looking into the network to the right of RB is
Vi
Zb re ( 1)RE
Ib
β
re
Zb
The result as displayed in Fig. 8.12 reveals that the input impedance of a transistor with an unbypassed resistor RE is determined by
Zb re ( 1)RE
RE
(8.20)
Since is normally much greater than 1, the approximate equation is the following:
Zb re RE
Figure 8.12 Defining the input
impedance of a transistor with an
unbypassed emitter resistor.
Zb (re RE)
and
(8.21)
Since RE is often much greater than re, Eq. (8.21) can be further reduced to
Zb RE
Zi:
(8.22)
Returning to Fig. 8.11, we have
Zi RBZb
(8.23)
Zo: With Vi set to zero, Ib 0 and Ib can be replaced by an open-circuit equivalent. The result is
Zo RC
(8.24)
Av :
Vi
Ib Zb
and
Vo Io RC IbRC
Vi
RC
Zb
with
346
Chapter 8
Vo
RC
Av Zb
Vi
BJT Small-Signal Analysis
(8.25)
Substituting Zb (re RE) gives
RC
Vo
Av re RE
Vi
(8.26)
and for the approximation Zb RE,
Vo
RC
Av Vi
RE
(8.27)
Note again the absence of from the equation for Av.
Ai: The magnitude of RB is often too close to Zb to permit the approximation
Ib Ii. Applying the current-divider rule to the input circuit will result in
RBIi
Ib RB Zb
and
Ib
RB
Ii
RB Zb
In addition,
Io Ib
and
Io
Ib
so that
Io
Io Ib
Ai Ii
Ib Ii
RB
RB Zb
and
or
Io
RB
Ai Ii
RB Zb
(8.28)
Zi
Ai Av
RC
(8.29)
Phase relationship: The negative sign in Eq. (8.25) again reveals a 180° phase
shift between Vo and Vi.
Effect of ro: The equations appearing below will clearly reveal the additional
complexity resulting from including ro in the analysis. Note in each case, however,
that when certain conditions are met, the equations return to the form just derived.
The derivation of each equation is beyond the needs of this text and is left as an exercise for the reader. Each equation can be derived through careful application of the
basic laws of circuit analysis such as Kirchhoff’s voltage and current laws, source
conversions, Thévenin’s theorem, and so on. The equations were included to remove
the nagging question of the effect of ro on the important parameters of a transistor
configuration.
Zi:
( 1) RC /ro
Zb re 1
(RC RE)/ro RE
(8.30)
8.4 CE Emitter-Bias Configuration
347
Since the ratio RC/ro is always much less than ( 1),
( 1)RE
Zb re 1 (RC RE)/ro
For ro 10(RC RE),
Zb re ( 1)RE
which compares directly with Eq. (8.20).
In other words, if ro 10(RC RE), all the equations derived earlier will result.
Since 1 , the following equation is an excellent one for most applications:
Zb (re RE)
Zo:
(8.31)
ro10(RCRE)
(ro re)
Zo RC ro re
1 RE
However, ro
(8.32)
re, and
Zo RC ro 1 re
1 RE
which can be written as
1
Zo RC ro 1 re
1
RE
Typically 1/ and re/RE are less than one with a sum usually less than one. The
result is a multiplying factor for ro greater than one. For 100, re 10 , and
RE 1 k:
1
1
1
50
re
1
10 1
0.02
RE
100
1000 Zo RC 51ro
and
which is certainly simply RC. Therefore,
Zo RC
(8.33)
Any level of ro
which was obtained earlier.
Av :
RC
RC
re
1 ro
Zb
ro
Vo
Av R
C
Vi
1 ro
348
Chapter 8
BJT Small-Signal Analysis
(8.34)
re
The ratio ro
1
RC
RC
Zb
ro
Vo RC
Av 1 Vi
ro
and
For ro 10RC,
Vo
RC
Av Zb
Vi
(8.35)
ro10RC
as obtained earlier.
Ai: The determination of Ai will be left to the equation
Zi
Ai Av
RC
(8.36)
using the above equations.
Bypassed
If RE of Fig. 8.10 is bypassed by an emitter capacitor CE, the complete re equivalent
model can be substituted resulting in the same equivalent network as Fig. 8.3. Eqs.
(8.1 through 8.9) are therefore applicable.
For the network of Fig. 8.13, without CE (unbypassed), determine:
(a) re.
(b) Zi.
20 V
(c) Zo.
(d) Av.
Io
(e) Ai.
EXAMPLE 8.3
2.2 kΩ
470 kΩ
10 µ F
Vo
C2
10 µ F
Zo
β = 120, ro = 40 kΩ
Vi
Ii
Zi
C1
0.56 kΩ
CE
10 µ F
Figure 8.13 Example 8.3.
Solution
VCC VBE
20 V 0.7 V
(a) DC: IB 35.89 A
470 k (121)0.56 k
RB ( 1)RE
IE ( 1)IB (121)(46.5 A) 4.34 mA
and
26 mV
26 mV
re 5.99 4.34 mA
IE
8.4 CE Emitter-Bias Configuration
349
(b) Testing the condition ro 10(RC RE),
40 k 10(2.2 k 0.56 k)
40 k 10(2.76 k) 27.6 k (satisfied )
Therefore,
Zb (re RE) 120(5.99 560 )
67.92 k
Zi RB Zb 470 k67.92 k
59.34 k
and
(c) Zo RC 2.2 k
(d) ro 10RC is satisfied. Therefore,
Vo
RC
(120)(2.2 k)
Av Zb
Vi
67.92 k
3.89
compared to 3.93 using Eq. (8.27): Av RC/RE.
Zi
59.34 k
(e) Ai Av (3.89) RC
2.2 k
104.92
compared to 104.85 using Eq. (8.28): Ai RB/(RB Zb).
EXAMPLE 8.4
Repeat the analysis of Example 8.3 with CE in place.
Solution
(a) The dc analysis is the same, and re 5.99 .
(b) RE is “shorted out” by CE for the ac analysis. Therefore,
Zi RBZb RBre 470 k(120)(5.99 )
470 k718.8 717.70 (c) Zo RC 2.2 k
RC
(d) Av re
2.2 k
367.28 (a significant increase)
5.99 RB
(120)(470 k)
(e) Ai 470
k 718.8 RB Zb
119.82
EXAMPLE 8.5
350
For the network of Fig. 8.14, determine (using appropriate approximations):
(a) re.
(b) Zi.
(c) Zo.
(d) Av.
(e) Ai.
Chapter 8
BJT Small-Signal Analysis
16 V
Io
2.2 kΩ
90 kΩ
C1
Ii
Zo
Zi
10 kΩ
0.68 kΩ
+
C2
β = 210, ro = 50 kΩ
Vi
Vo
CE
–
Figure 8.14 Example 8.5.
Solution
(a) Testing RE
10R2
(210)(0.68 k)
142.8 k
10(10 k)
100 k (satisfied)
R2
10 k
VB VCC (16 V) 1.6 V
R1 R2
90 k 10 k
VE VB VBE 1.6 V 0.7 V 0.9 V
VE
0.9 V
IE 1.324 mA
RE
0.68 k
26 mV
26 mV
re 19.64 IE
1.324 mA
(b) The ac equivalent circuit is provided in Fig. 8.15. The resulting configuration is
now different from Fig. 8.11 only by the fact that now
RB R R1R2 9 k
Ii
+
Io
+
Vi
10 kΩ
90 kΩ
2.2 kΩ
Vo
0.68 kΩ
–
Zi
Zo
–
R'
Figure 8.15 The ac equivalent circuit of Fig. 8.14.
The testing conditions of ro 10 (RC RE) and ro 10RC are both satisfied. Using
the appropriate approximations yields
Zb RE 142.8 k
Zi RBZb 9 k142.8 k
8.47 k
8.4 CE Emitter-Bias Configuration
351
(c) Zo RC 2.2 k
2.2 k
RC
(d) Av 3.24
0.68 k
RE
Zi
8.47 k
(e) Ai Av (3.24) RC
2.2 k
12.47
EXAMPLE 8.6
Repeat Example 8.5 with CE in place.
Solution
(a) The dc analysis is the same, and re 19.64 .
(b) Zb re (210)(19.64 ) 4.12 k
Zi RBZb 9 k4.12 k
2.83 k
(c) Zo RC 2.2 k
RC
2.2 k
(d) Av 112.02 (a significant increase)
re
19.64 k
Zi
2.83 k
(e) Ai Av (112.02) RL
2.2 k
144.1
Another variation of an emitter-bias configuration appears in Fig. 8.16. For the dc
analysis, the emitter resistance is RE1 RE2, while for the ac analysis, the resistor RE
in the equations above is simply RE1 with RE2 bypassed by CE.
VCC
Io
RC
C2
RB
Vo
C1
Vi
Ii
Zo
RE
1
Zi
RE
2
CE
Figure 8.16 An emitter-bias
configuration with a portion of
the emitter-bias resistance bypassed in the ac domain.
8.5 EMITTER-FOLLOWER
CONFIGURATION
When the output is taken from the emitter terminal of the transistor as shown in Fig.
8.17, the network is referred to as an emitter-follower. The output voltage is always
slightly less than the input signal due to the drop from base to emitter, but the ap352
Chapter 8
BJT Small-Signal Analysis
VCC
C
RB
Ii
B
Vi
C1
C2
Vo
E
Io
Zi
RE
Zo
Figure 8.17 Emitter-follower
configuration.
proximation Av 1 is usually a good one. Unlike the collector voltage, the emitter
voltage is in phase with the signal Vi. That is, both Vo and Vi will attain their positive and negative peak values at the same time. The fact that Vo “follows” the magnitude of Vi with an in-phase relationship accounts for the terminology emitterfollower.
The most common emitter-follower configuration appears in Fig. 8.17. In fact, because the collector is grounded for ac analysis, it is actually a common-collector configuration. Other variations of Fig. 8.17 that draw the output off the emitter with Vo Vi will appear later in this section.
The emitter-follower configuration is frequently used for impedance-matching purposes. It presents a high impedance at the input and a low impedance at the output,
which is the direct opposite of the standard fixed-bias configuration. The resulting effect is much the same as that obtained with a transformer, where a load is matched
to the source impedance for maximum power transfer through the system.
Substituting the re equivalent circuit into the network of Fig. 8.17 will result in
the network of Fig. 8.18. The effect of ro will be examined later in the section.
Ii
b
c
Ib
+
β Ib
β re
Zi
Vi
RB
e
Zb
–
RE
+
Io
Zo
Ie = ( β + 1) Ib
Vo
Figure 8.18 Substituting the re
equivalent circuit into the ac
equivalent network of Fig. 8.17.
–
Zi: The input impedance is determined in the same manner as described in the
preceding section:
with
Zi RBZb
(8.37)
Zb re ( 1)RE
(8.38)
8.5 Emitter-Follower Configuration
353
or
Zb (re RE)
(8.39)
Zb RE
(8.40)
and
Zo: The output impedance is best described by first writing the equation for the
current Ib:
Vi
Ib Zb
and then multiplying by ( 1) to establish Ie. That is,
Vi
Ie ( 1)Ib ( 1) Zb
Substituting for Zb gives
( 1)Vi
Ie re ( 1)RE
or
Vi
Ie [re/( 1)] RE
but
( 1) and
re
re
re
1
Vi
Ie re RE
so that
re
+
Vo
Ie
(8.41)
If we now construct the network defined by Eq. (8.41), the configuration of Fig.
8.19 will result.
To determine Zo, Vi is set to zero and
RE
Vi
–
Zo
Figure 8.19 Defining the output impedance for the emitter-follower configuration.
Zo REre
(8.42)
Since RE is typically much greater than re, the following approximation is often applied:
Zo re
(8.43)
Av: Figure 8.19 can be utilized to determine the voltage gain through an application of the voltage-divider rule:
REVi
Vo RE re
and
Vo
RE
Av Vi
RE re
(8.44)
Since RE is usually much greater than re, RE re RE and
Vo
Av 1
Vi
354
Chapter 8
BJT Small-Signal Analysis
(8.45)
Ai:
From Fig. 8.18,
RBIi
Ib RB Zb
or
Ib
RB
Ii
RB Zb
and
Io Ie ( 1)Ib
or
Io
( 1)
Ib
Io
Io Ib
Ai Ii
Ib Ii
so that
RB
( 1) RB Zb
( 1) ,
and since
or
RB
Ai RB Zb
(8.46)
Zi
Ai Av RE
(8.47)
Phase relationship: As revealed by Eq. (8.44) and earlier discussions of this
section, Vo and Vi are in phase for the emitter-follower configuration.
Effect of ro:
Zi:
( 1)RE
(8.48)
Zb re RE
1 ro
If the condition ro 10RE is satisfied,
Zb re ( 1)RE
which matches earlier conclusions with
Zb (re RE)
Zo:
ro10RE
re
Zo ro RE ( 1)
(8.49)
(8.50)
Using 1 ,
Zo roREre
and since ro
re,
Zo RE re
(8.51)
Any ro
8.5 Emitter-Follower Configuration
355
Av:
( 1)RE/Zb
Av RE
1 ro
(8.52)
If the condition ro 10RE is satisfied and we use the approximation 1 ,
RE
Av Zb
EXAMPLE 8.7
But
Zb (re RE)
so that
RE
Av (re RE)
and
RE
Av re RE
(8.53)
ro10RE
For the emitter-follower network of Fig. 8.20, determine:
(a) re.
(b) Zi.
(c) Zo.
(d) Av.
(e) Ai.
(f) Repeat parts (b) through (e) with ro 25 k and compare results.
12 V
220 kΩ
10 µ F
β = 100, ro = ∞ Ω
Vi
Ii
10 µ F
Vo
Io
Zi
3.3 kΩ
Zo
Figure 8.20 Example 8.7.
Solution
VCC VBE
(a) IB RB ( 1)RE
12 V 0.7 V
20.42 A
220 k (101)3.3 k
IE ( 1)IB
(101)(20.42 A) 2.062 mA
26 mV
26 mV
re 12.61 2.062 mA
IE
356
Chapter 8
BJT Small-Signal Analysis
(b) Zb re ( 1)RE
(100)(12.61 ) (101)(3.3 k)
1.261 k 333.3 k
334.56 k RE
Zi RB Zb 220 k334.56 k
132.72 k
(c) Zo REre 3.3 k12.61 12.56 re
3.3 k
Vo
RE
(d) Av RE re
3.3 k 12.61 Vi
0.996 1
(100)(220 k)
RB
(e) Ai 39.67
220 k 334.56 k
RB Zb
versus
Zi
132.72 k
Ai Av (0.996) 40.06
RE
3.3 k
(f) Checking the condition ro 10RE, we have
25 k 10(3.3 k) 33 k
which is not satisfied. Therefore,
( 1)RE
(100 1)3.3 k
(100)(12.61 ) Zb re RE
3.3 k
1 1 ro
25 k
with
1.261 k 294.43 k
295.7 k
Zi RBZb 220 k295.7 k
126.15 k vs. 132.72 k obtained earlier
Zo RE re 12.56 as obtained earlier
( 1)RE/Zb
(100 1)(3.3 k)/295.7 k
Av RE
3.3 k
1 1
25 k
ro
0.996 1
matching the earlier result.
In general, therefore, even though the condition ro 10RE was not satisfied, the
results for Zo and Av are the same, with Zi only slightly less. The results suggest that
for most applications a good approximation for the actual results can be obtained by
simply ignoring the effects of ro for this configuration.
The network of Fig. 8.21 is a variation of the network of Fig. 8.17, which employs a voltage-divider input section to set the bias conditions. Equations (8.37)
through (8.47) are changed only by replacing RB by R R1R2.
The network of Fig. 8.22 will also provide the input/output characteristics of an
emitter-follower but includes a collector resistor RC. In this case RB is again replaced
by the parallel combination of R1 and R2. The input impedance Zi and output impedance Zo are unaffected by RC since it is not reflected into the base or emitter equiv8.5 Emitter-Follower Configuration
357
VCC
VCC
R1
RC
R1
Ii
C1
Vi
Vi
C2
C1
C2
Vo
R2
Zi
Vo
Io
R2
Zi
RE
Io
RE
Zo
Zo
Figure 8.22 Emitter-follower
configuration with a collector
resistor RC.
Figure 8.21 Emitter-follower
configuration with a voltagedivider biasing arrangement.
alent networks. In fact, the only effect of RC will be to determine the Q-point of operation.
8.6 COMMON-BASE CONFIGURATION
The common-base configuration is characterized as having a relatively low input and
a high output impedance and a current gain less than 1. The voltage gain, however,
can be quite large. The standard configuration appears in Fig. 8.23, with the commonbase re equivalent model substituted in Fig. 8.24. The transistor output impedance ro
is not included for the common-base configuration because it is typically in the
megohm range and can be ignored in parallel with the resistor RC.
Ii
Ie
Ic
+
Vi
–
E
Zi
C
RE
+
Io
RC
Vo
B
VEE
e
VCC
–
+
Zo
Ic
Ii
Vi
–
Ie
c
Io
RE
re
α Ie
RC
Zi
+
Vo
Zo
–
Figure 8.23 Common-base configuration.
Figure 8.24 Substituting the re equivalent circuit into the
ac equivalent network of Fig. 8.23.
Zi:
Zi RE re
(8.54)
Zo RC
(8.55)
Zo:
Av :
Vo Io RC (Ic )RC IeRC
358
Chapter 8
BJT Small-Signal Analysis
Vi
Ie re
with
Vo so that
r R
Vi
C
e
RC
Vo
RC
Av re
Vi
re
and
A i:
Assuming that RE
(8.56)
re yields
Ie Ii
and
Io Ie Ii
with
Io
Ai 1
Ii
(8.57)
Phase relationship: The fact that Av is a positive number reveals that Vo and Vi
are in phase for the common-base configuration.
Effect of ro: For the common-base configuration, ro 1/hob is typically in the
megohm range and sufficiently larger than the parallel resistance RC to permit the approximation roRC RC.
For
(a)
(b)
(c)
(d)
(e)
the network of Fig. 8.25, determine:
re.
Zi.
Zo.
Av.
Ai.
10 µ F
+
Ii
Zi
10 µF
Ie
1 kΩ
Vo
EXAMPLE 8.8
α = 0.98
ro = 1 MΩ
+
Io
5 kΩ
2V
8V
–
Zo
Vo
–
Figure 8.25 Example 8.8.
Solution
VEE VBE
2 V 0.7 V
1.3 V
(a) IE 1.3 mA
RE
1 k
1 k
(b)
(c)
(d)
(e)
26 mV
26 mV
re 20 IE
1.3 mA
Zi REre 1 k20 19.61 re
Zo RC 5 k
RC
5 k
Av 250
re
20 Ai 0.98 1
8.6 Common-Base Configuration
359
8.7 COLLECTOR FEEDBACK
CONFIGURATION
The collector feedback network of Fig. 8.26 employs a feedback path from collector
to base to increase the stability of the system as discussed in Section 4.12. However,
the simple maneuver of connecting a resistor from base to collector rather than base
to dc supply has a significant impact on the level of difficulty encountered when analyzing the network.
VCC
RC
RF
Io
C
Vo
C2
Ii
B
Vi
Zo
C1
E
Figure 8.26 Collector feedback
configuration.
Zi
Some of the steps to be performed below are the result of experience working
with such configurations. It is not expected that a new student of the subject would
choose the sequence of steps described below without taking a wrong step or two.
Substituting the equivalent circuit and redrawing the network will result in the configuration of Fig. 8.27. The effects of a transistor output resistance ro will be discussed later in the section.
B
+
Ii
–
Ib
β re
Vi
RF
I'
+
Io
C
+
Ic
β Ib
RC
Zi
–
Zo V
o
–
Figure 8.27 Substituting the re
equivalent circuit into the ac
equivalent network of Fig. 8.26.
Vo Vi
I RF
Zi:
with
Vo IoRC
and
Io Ib I
Since Ib is normally much larger than I ,
Io Ib
360
and
Vo (Ib)RC IbRC
but
Vi
Ib re
and
Vi
RC
Vo RC Vi
re
re
Chapter 8
BJT Small-Signal Analysis
Therefore,
Vo
Vi
Vi
1
Vo Vi
RCVi
RC
I 1 Vi
RF
RF RF
RF
RF
reRF
re
The result is
Vi Ibre (Ii I )re Iire I re
1
RC
Vi Iire 1 reVi
RF
re
re
RC
Vi 1 1 Iire
RF
re
or
Vi
re
Zi re
RC
Ii
1 1 r
RF
RC eRC
but RC is usually much greater than re and 1 re
re
and
re
Zi RC
1 RF
so that
re
Zi 1
RC
RF
or
(8.58)
Zo: If we set Vi to zero as required to define Zo, the network will appear as
shown in Fig. 8.28. The effect of re is removed and RF appears in parallel with RC
and
Zo RC RF
(8.59)
RF
Ib = 0 A
Vi = 0
β re
β Ib = 0 A
RC
Zo
Figure 8.28 Defining Zo for the
collector feedback configuration.
Av:
At node C of Fig. 8.27,
Io Ib I
For typical values, Ib
I and Io Ib.
Vo Io RC (Ib)RC
Substituting Ib Vi/re gives us
Vi
Vo RC
re
and
Vo
RC
Av Vi
re
(8.60)
8.7 Collector Feedback Configuration
361
Ai:
Applying Kirchhoff’s voltage law around the outside network loop yields
Vi VRF Vo 0
Ibre (Ib Ii)RF IoRC 0
and
Using Io Ib, we have
Ibre Ib RF Ii RF IbRC 0
Ib(re RF RC) IiRF
and
Substituting Ib Io/ from Io Ib yields
Io
(re RF RC) IiRF
RFIi
Io re RF RC
and
Ignoring re compared to RF and RC gives us
Io
RF
Ai Ii
RF RC
For RC
(8.61)
RF,
Io
RF
Ai Ii
RC
Io
RF
Ai Ii
RC
and
(8.62)
Phase relationship: The negative sign of Eq. (8.60) reveals a 180° phase shift
between Vo and Vi.
Effect of ro:
Zi: A complete analysis without applying approximations will result in
RC ro
1 RF
Zi 1
1
RC ro
re
RF
RFre
Recognizing that 1/RF 0 and applying the condition ro 10RC,
RC
1 RF
RC
Zi 1
re
RFre
but typically RC/RF
1 and
1
Zi RC
1
re
RFre
362
Chapter 8
BJT Small-Signal Analysis
(8.63)
re
Zi 1
RC
RF
or
(8.64)
ro10RC
as obtained earlier.
Zo: Including ro in parallel with RC in Fig. 8.28 will result in
Zo roRC RF
(8.65)
For ro 10RC,
Zo RC RF
(8.66)
ro10RC
as obtained earlier. For the common condition of RF
Zo RC
RC,
(8.67)
ro10RC, RF
RC
Av :
R r(r R )
1
1
F
o
e
C
Av roRC
1
RF
Since RF
re,
(8.68)
roRC
re
roRC
Av 1 RF
For ro 10RC,
RC
re
RC
Av 1 RF
(8.69)
ro10RC
and since RC/RF is typically much less than one,
RC
Av re
(8.70)
ro10RC, RF
RC
as obtained earlier.
For
(a)
(b)
(c)
(d)
(e)
(f)
the network of Fig. 8.29, determine:
re.
Zi.
Zo.
Av.
Ai.
Repeat parts (b) through (e) with ro 20 k and compare results.
8.7 Collector Feedback Configuration
EXAMPLE 8.9
363
9V
2.7 kΩ
180 kΩ
Io
Vo
10 µF
Ii
Vi
10 µF
β = 200, ro = ∞ Ω
Zo
Zi
Figure 8.29 Example 8.9.
Solution
9 V 0.7 V
VCC VBE
(a) IB 180
k
(200)2.7 k
RF RC
11.53 A
IE ( 1)IB (201)(11.53 A) 2.32 mA
26 mV
26 mV
re 11.21 2.32 mA
IE
11.21 re
11.21 (b) Zi 0.005 0.015
1
RC
1
2.7 k
200
180 k
RF
11.21 50(11.21 ) 560.5 0.02
(c) Zo RCRF 2.7 k180 k 2.66 k
27 k
RC
(d) Av 240.86
11.21 re
(200)(180 k)
RF
(e) Ai 180 k (200)(2.7 k)
RF RC
50
(f) Zi: The condition
R Cro ro 10RC is not satisfied. Therefore,
2.7 k20 k
1 1 RF
180 k
1
RC ro Zi 1
1
1
02.7 k 20 k0
re
RF
RFre
(200)(11.21)
180 k
(180 k)(11.21 )
2.38 k
1 180 k
1 0.013
3
0.45 10 0.006 103 1.18 103 1.64 103
617.7 vs. 560.5 above
Zo:
Zo roRCRF 20 k2.7 k180 k
2.35 k vs. 2.66 k above
364
Chapter 8
BJT Small-Signal Analysis
Av:
1
1
1
1
(roRC)
(2.38 k)
RF re
180 k
11.21
Av roRC
2.38 k
1
1 RF
180 k [5.56 106 8.92 102](2.38 k)
1 0.013
209.56 vs. 240.86 above
Ai:
Zi
Ai Av RC
617.7 (209.56)
2.7 k
47.94 vs. 50 above
For the configuration of Fig. 8.30, Eqs. (8.71) through (8.74) will determine the
variables of interest. The derivations are left as an exercise at the end of the chapter.
VCC
RC
Io
RF
Vo
Ii
C2
Vi
C1
Zo
Zi
RE
Figure 8.30 Collector feedback
configuration with an emitter resistor RE.
Zi:
RE
Zi 1
(RE RC)
RF
(8.71)
Zo:
Zo RCRF
(8.72)
RC
Av RE
(8.73)
Av :
A i:
1
Ai 1
(RE RC)
RF
(8.74)
8.7 Collector Feedback Configuration
365
8.8 COLLECTOR DC FEEDBACK
CONFIGURATION
The network of Fig. 8.31 has a dc feedback resistor for increased stability, yet the capacitor C3 will shift portions of the feedback resistance to the input and output sections of the network in the ac domain. The portion of RF shifted to the input or output side will be determined by the desired ac input and output resistance levels.
VCC
RC
RF
RF
1
Io
2
Vo
C2
C3
C1
Vi
Zo
Ii
Figure 8.31 Collector dc feedback configuration.
Zi
At the frequency or frequencies of operation, the capacitor will assume a shortcircuit equivalent to ground due to its low impedance level compared to the other elements of the network. The small-signal ac equivalent circuit will then appear as
shown in Fig. 8.32.
Ii
Ib
+
Zi
Vi
+
Io
RF
1
β re
β Ib
ro
RF
2
RC
Vo
Zo
–
–
Figure 8.32 Substituting the re
equivalent circuit into the ac
equivalent network of Fig. 8.31.
Zi:
Zi RF1re
(8.75)
Zo RC RF2ro
(8.76)
Zo:
For ro 10RC,
Zo RC RF2
(8.77)
ro10RC
Av :
R ro RF2RC
Vo Ib R
and
366
Chapter 8
BJT Small-Signal Analysis
but
Vi
Ib re
and
Vi
Vo R
re
so that
roRF2RC
Vo
Av Vi
re
(8.78)
For ro 10RC,
RF RC
Vo
Av 2 re
Vi
(8.79)
ro10RC
Ai: For the input side,
RF1Ii
Ib RF1 re
or
RF1
Ib
RF1 re
Ii
and for the output side using R roRF2
R Ib
Io R RC
or
R
Io
R RC
Ib
The current gain,
Io
Io Ib
Ai Ii
Ib Ii
RF1
R
R RC RF1 re
and
RF1R
Io
Ai (R
re)(R RC)
Ii
F1
(8.80)
R roRF2
Since RF1 is usually much larger than re, RF1 re RF1
and
so that
Io
RF1(roRF2)
Ai Ii
RF1(roRF2 RC)
Io
Ai Ii
RC
1 roRF2
(8.81)
RF110re
or
Io
Zi
Ai Av
Ii
RC
(8.82)
Phase relationship: The negative sign in Eq. (8.78) clearly reveals a 180° phase
shift between input and output voltages.
8.8 Collector DC Feedback Configuration
367
EXAMPLE 8.10
For
(a)
(b)
(c)
(d)
(e)
the network of Fig. 8.33, determine:
re.
12 V
Zi.
Zo.
Av.
3 kΩ
Ai.
120 kΩ
68 kΩ
Io
Vo
0.01 µF
10 µF
Ii
Zo
β = 140, ro = 30 kΩ
Vi
10 µF
Zi
Figure 8.33 Example 8.10.
Solution
(a)
DC:
VCC VBE
IB RF RC
12 V 0.7 V
(120 k 68 k) (140)3 k
11.3V
18.6 A
608 k
IE ( 1)IB (141)(18.6 A)
2.62 mA
26 mV
26 mV
re 9.92 IE
2.62 mA
(b) re (140)(9.92 ) 1.39 k
The ac equivalent network appears in Fig. 8.34.
Zi RF1re 120 k1.39 k
1.37 k
+
Vi
Ib
120 kΩ
+
Io
Ii
β re
1.395 kΩ
β Ib
140 Ib
ro
30 kΩ
68 kΩ
Zi
3 kΩ
Vo
Zo
–
Figure 8.34 Substituting the re equivalent circuit into the ac equivalent network of
Fig. 8.33.
(c) Testing the condition ro 10RC, we find
30 k 10(3 k) 30 k
which is satisfied through the equals sign in the condition. Therefore,
Zo RCRF2 3 k68 k
2.87 k
368
Chapter 8
BJT Small-Signal Analysis
–
(d) ro 10RC, therefore,
RF RC
68 k3 k
Av 2 re
9.92 2.87 k
9.92 289.3
re is satisfied,
(e) Since the condition RF1
140
140
140
Ai 1.14
1 0.14
RC
3 k
1 1 roRF2
30 k68 k
122.8
8.9 APPROXIMATE HYBRID EQUIVALENT
CIRCUIT
The analysis using the approximate hybrid equivalent circuit of Fig. 8.35 for the
common-emitter configuration and of Fig. 8.36 for the common-base configuration is
very similar to that just performed using the re model. Although time and priorities
do not permit a detailed analysis of all the configurations discussed thus far, a brief
overview of some of the most important will be included in this section to demonstrate the similarities in approach and the resulting equations.
b
c
Ib
hie
hfe Ib
hoe
e
e
e
Figure 8.35 Approximate
common-emitter hybrid equivalent circuit.
c
Ie
hib
b
hfb Ie
hob
b
Figure 8.36 Approximate
common-base hybrid equivalent
circuit.
Since the various parameters of the hybrid model are specified by a data sheet or
experimental analysis, the dc analysis associated with use of the re model is not an
integral part of the use of the hybrid parameters. In other words, when the problem
is presented, the parameters such as hie, hfe, hib, and so on, are specified. Keep in
mind, however, that the hybrid parameters and components of the re model are related by the following equations as discussed in detail in Chapter 7: hie re, hfe , hoe 1/ro, hfb , and hib re (note Appendix A).
Fixed-Bias Configuration
For the fixed-bias configuration of Fig. 8.37, the small-signal ac equivalent network
will appear as shown in Fig. 8.38 using the approximate common-emitter hybrid equiv8.9 Approximate Hybrid Equivalent Circuit
369
VCC
RC
Io
RB
hie
hfe
+
C1
Vi
+
Ii
C2
Zo
Vo
Zi
–
–
Ic
RB
Io
hie
hfe Ib
hoe
RC
Vo
Zo
–
Ib
Zi
+
Ii
+
Vi
Figure 8.37 Fixed-bias configuration.
–
Figure 8.38 Substituting the approximate hybrid equivalent circuit into
the ac equivalent network of Fig. 8.37.
alent model. Compare the similarities in appearance with Fig. 8.3 and the re model
analysis. The similarities suggest that the analysis will be quite similar, and the results of one can be directly related to the other.
Zi: From Fig. 8.38,
Zo:
Av :
Zi RBhie
(8.83)
Zo RC1/hoe
(8.84)
From Fig. 8.38,
Using R 1/hoeRC,
Vo Io R ICR
hfe Ib R
and
Vi
Ib hie
with
Vi
Vo hfe R
hie
hfe(RC1/hoe)
Vo
Av hie
Vi
so that
Ai: Assuming that RB
hfeIb hfe Ii with
hie and 1/hoe 10RC, then Ib Ii and Io Ic Io
Ai hfe
Ii
370
Chapter 8
(8.85)
BJT Small-Signal Analysis
(8.86)
EXAMPLE 8.11
For the network of Fig. 8.39, determine:
(a)
(b)
(c)
(d)
8V
Io
2.7 kΩ
330 kΩ
Zi.
Zo.
Av.
Ai.
Vo
Vi
hfe = 120
Zo
hie = 1.175 kΩ
hoe = 20 µ A/V
Ii
Figure 8.39 Example 8.11.
Zi
Solution
Zi RBhie 330 k1.175 k
hie 1.171 k
1
1
(b) ro 50 k
hoe
20 A/V
1
Zo RC 50 k2.7 k 2.56 k RC
hoe
(120)(2.7 k50 k)
hfe(RC1/hoe)
(c) Av 262.34
1.171 k
hie
(d) Ai hfe 120
(a)
Voltage-Divider Configuration
For the voltage-divider bias configuration of Fig. 8.40, the resulting small-signal ac
equivalent network will have the same appearance as Fig. 8.38, with RB replaced by
R R1R2.
VCC
Io
RC
R1
Vo
Ii
C2
hie
hfe
Vi
C1
Zo
R2
RE
Zi
CE
Figure 8.40 Voltage-divider
bias configuration.
Zi:
Zo:
From Fig. 8.38 with RB R ,
Zi R hie
(8.87)
Zo RC
(8.88)
From Fig. 8.38,
8.9 Approximate Hybrid Equivalent Circuit
371
Av:
hfe(RC1/hoe)
Av hie
(8.89)
hf R
Ai e
R hie
(8.90)
A i:
Unbypassed Emitter-Bias Configuration
For the CE unbypassed emitter-bias configuration of Fig. 8.41, the small-signal ac
model will be the same as Fig. 8.11, with re replaced by hie and Ib by hfeIb. The
analysis will proceed in the same manner.
VCC
Io
RC
RB
Vo
hie
hfe
Vi
Ii
Zo
RE
Zi
Figure 8.41 CE unbypassed
emitter-bias configuration.
Zi:
and
Zb hfe RE
(8.91)
Zi RBZb
(8.92)
Zo RC
(8.93)
Zo:
Av :
hfe RC
hfeRC
Av Zb
hfe RE
and
RC
Av RE
(8.94)
hf RB
Ai e
RB Zb
(8.95)
A i:
372
Chapter 8
BJT Small-Signal Analysis
Zi
Ai Av
RC
or
(8.96)
Emitter-Follower Configuration
For the emitter-follower of Fig. 8.42, the small-signal ac model will match Fig. 8.18,
with re hie and hfe. The resulting equations will therefore be quite similar.
Zi:
Zb hfe RE
(8.97)
Zi RBZb
(8.98)
VCC
RB
Ii
Vi
hie
hfe
Zi
Vo
Io
Zo
RE
Figure 8.42 Emitter-follower
configuration.
Zo: For Zo, the output network defined by the resulting equations will appear
as shown in Fig. 8.43. Review the development of the equations in Section 8.5 and
hie
Zo RE 1 hfe
or since 1 hfe hfe,
hie
Zo RE hfe
(8.99)
Figure 8.43 Defining Zo for the
emitter-follower configuration.
Av: For the voltage gain, the voltage-divider rule can be applied to Fig. 8.43 as
follows:
RE (Vi)
Vo RE hie/(1 hfe)
8.9 Approximate Hybrid Equivalent Circuit
373
but since 1 hfe hfe,
RE
Vo
Av RE hie/hfe
Vi
(8.100)
hfe RB
Ai RB Zb
(8.101)
Zi
Ai Av
RE
(8.102)
A i:
or
Common-Base Configuration
The last configuration to be examined with the approximate hybrid equivalent circuit
will be the common-base amplifier of Fig. 8.44. Substituting the approximate common-base hybrid equivalent model will result in the network of Fig. 8.45, which is
very similar to Fig. 8.24. From Fig. 8.45,
hib , hfb
Ii
Ic
+
Vi
Io
RE
Zi
+
RC
Vo
VEE
Zo
VCC
–
–
Figure 8.44 Common-base configuration.
+
Zi
Vi
Ie
Ii
RE
Io
hib
hfb Ie
RC
–
+
Vo
Zo
–
Figure 8.45 Substituting the approximate hybrid equivalent circuit into the ac
equivalent network of Fig. 8.44.
Zi:
Zi RE hib
(8.103)
Zo RC
(8.104)
Zo:
Av :
Vo Io RC (hf b Ie)RC
374
Chapter 8
BJT Small-Signal Analysis
Vi
Ie hib
with
so that
and
Vi
Vo hfb RC
hib
Vo
hfb RC
Av hib
Vi
(8.105)
Io
Ai hfb 1
Ii
(8.106)
A i:
For
(a)
(b)
(c)
(d)
the network of Fig. 8.46, determine:
Zi.
Zo.
Av.
Ai.
EXAMPLE 8.12
Ii
+
Vi
Io
Zi
2.2 kΩ
4V
hfb = − 0.99
hib = 14.3 Ω
hob = 0.5 µA/V
+
3.3 kΩ
Vo
Zo
10 V
–
–
Figure 8.46 Example 8.12.
Solution
(a) Zi REhib 2.2 k14.3 14.21 hib
1
1
(b) ro 2 M
hob
0.5 A/V
1
Zo RC RC 3.3 k
hob
hfb RC
(0.99)(3.3 k)
(c) Av 229.91
hib
14.21
(d) Ai hfb 1
The remaining configurations of Sections 8.1 through 8.8 that were not analyzed
in this section are left as an exercise in the problem section of this chapter. It is assumed that the analysis above clearly reveals the similarities in approach using the re
or approximate hybrid equivalent models, thereby removing any real difficulty with
analyzing the remaining networks of the earlier sections.
8.10 COMPLETE HYBRID EQUIVALENT
MODEL
The analysis of Section 8.9 was limited to the approximate hybrid equivalent circuit
with some discussion about the output impedance. In this section, we employ the complete equivalent circuit to show the impact of hr and define in more specific terms the
8.10 Complete Hybrid Equivalent Model
375
impact of ho. It is important to realize that since the hybrid equivalent model has the
same appearance for the common-base, common-emitter, and common-collector configurations, the equations developed in this section can be applied to each configuration. It is only necessary to insert the parameters defined for each configuration. That
is, for a common-base configuration, hfb, hib, and so on, are employed, while for a
common-emitter configuration, hfe, hie, and so on, are utilized. Recall that Appendix
A permits a conversion from one set to the other if one set is provided and the other
is required.
Consider the general configuration of Fig. 8.47 with the two-port parameters of
particular interest. The complete hybrid equivalent model is then substituted in Fig.
8.48 using parameters that do not specify the type of configuration. In other words,
the solutions will be in terms of hi, hr, hf, and ho. Unlike the analysis of previous sections of this chapter, the current gain Ai will be determined first since the equations
developed will prove useful in the determination of the other parameters.
Io
+
Ii
Rs
+
Vi
Zi
Vs
+
–
Transistor
–
Zo
Vo
RL
–
Figure 8.47 Two-port system.
Ii
+
Ib
hi
Rs
+
Vs
–
Zi
Vi
I
+
1/ho
Vo
+
hr Vo
hf Ib
–
–
Io
Zo
–
Figure 8.48 Substituting the complete hybrid equivalent circuit into the two-port system of Fig. 8.47.
Current Gain, Ai Io /Ii
Applying Kirchhoff’s current law to the output circuit yields
Vo
Io hf Ib I hf Ii hf Ii hoVo
1/ho
Substituting Vo Io RL gives us
Io hf Ii ho RL Io
Rewriting the equation above, we have
Io ho RL Io hf Ii
and
376
Chapter 8
Io(1 ho RL) hf Ii
BJT Small-Signal Analysis
RL
Io
hf
Ai Ii
1 hoRL
so that
(8.107)
Note that the current gain will reduce to the familiar result of Ai hf if the factor
hoRL is sufficiently small compared to 1.
Voltage Gain, Av Vo /Vi
Applying Kirchhoff’s voltage law to the input circuit results in
Vi Iihi hrVo
Substituting Ii (1 hoRL)Io/hf from Eq. (8.107) and Io Vo/RL from above results in
(1 hoRL)hi
Vo hrVo
Vi hf RL
Solving for the ratio Vo/Vi yields
hf RL
Vo
Av hi (hiho hf hr)RL
Vi
(8.108)
In this case, the familiar form of Av hf RL/hi will return if the factor (hiho hf hr)RL is sufficiently small compared to hi.
Input Impedance, Zi Vi/Ii
For the input circuit,
Vi hi Ii hrVo
Substituting
Vo Io RL
we have
Vi hi Ii h r RL Io
Since
Io
Ai Ii
Io Ai Ii
so that the equation above becomes
Vi hiIi hr RL Ai Ii
Solving for the ratio Vi/Ii, we obtain
Vi
Zi hi hr RL Ai
Ii
and substituting
hf
Ai 1 ho RL
yields
Vi
h f hr RL
Zi hi Ii
1 ho RL
(8.109)
The familiar form of Zi hi will be obtained if the second factor is sufficiently smaller
than the first.
8.10 Complete Hybrid Equivalent Model
377
Output Impedance, Zo Vo /Io
The output impedance of an amplifier is defined to be the ratio of the output voltage
to the output current with the signal Vs set to zero. For the input circuit with Vs 0,
hrVo
Ii Rs hi
Substituting this relationship into the following equation obtained from the output circuit yields
Io hf Ii hoVo
hf hrVo
hoVo
Rs hi
Vo
1
Zo Io
ho [hf hr /(hi Rs)]
and
(8.110)
In this case, the output impedance will reduce to the familiar form Zo 1/ho for the
transistor when the second factor in the denominator is sufficiently smaller than the
first.
EXAMPLE 8.13
For the network of Fig. 8.49, determine the following parameters using the complete
hybrid equivalent model and compare to the results obtained using the approximate
model.
(a) Zi and Z i.
(b) Av.
(c) Ai Io /Ii and A i Io /I i.
(d) Zo (within RC) and
Z o (including RC).
Figure 8.49 Example 8.13.
Solution
Now that the basic equations for each quantity have been derived, the order in which
they are calculated is arbitrary. However, the input impedance is often a useful quantity to know and therefore will be calculated first. The complete common-emitter hybrid equivalent circuit has been substituted and the network redrawn as shown in Fig.
8.50. A Thévenin equivalent circuit for the input section of Fig. 8.50 will result in the
input equivalent of Fig. 8.51 since ETh Vs and RTh Rs 1 k (a result of RB 470 k being much greater than Rs 1 k). In this example, RL RC and Io is de378
Chapter 8
BJT Small-Signal Analysis
Ii'
Rs
1 kΩ
Io
Ii
Ib
+
Z'i
+
Zi
1.6 kΩ
+
2 × 10− 4 Vo
470 kΩ
Vi
Z'o
Zo
110 Ib
50 kΩ
4.7 kΩ
+
Vo
–
Vs
–
–
–
Thevenin
Figure 8.50 Substituting the complete hybrid equivalent circuit into the ac
equivalent network of Fig. 8.49.
I'i
Rs
1 kΩ
Ii
Z'i
+
+
Zi
Io
hie
1.6 kΩ
Z'o
+
hre Vo
2 × 10− 4 Vo
Vi
hfe Ib
110 Ib
–
Vs
–
1
hoe = 50 kΩ
hoe = 20 µS
–
+
Zo
4.7 kΩ
Vo
–
Figure 8.51 Replacing the input section of Fig. 8.50 with a Thévenin equivalent circuit.
fined as the current through RC as in previous examples of this chapter. The output
impedance Zo as defined by Eq. (8.110) is for the output transistor terminals only. It
does not include the effects of RC. Z o is simply the parallel combination of Zo and
RL. The resulting configuration of Fig. 8.51 is then an exact duplicate of the defining
network of Fig. 8.48, and the equations derived above can be applied.
Vi
h fe h re RL
(a) Eq. (8.109): Zi hie Ii
1 hoe RL
(110)(2 104)(4.7 k)
1.6 k 1 (20 S)(4.7 k)
1.6 k 94.52 1.51 k
versus 1.6 k using simply hie.
Z i 470 kZi Zi 1.51 k
(b) Eq. (8.108):
hfeRL
Vo
Av hie (hiehoe hfehre)RL
Vi
(110)(4.7 k)
1.6 k [(1.6 k)(20 S) (110)(2 104)]4.7 k
517 103 1.6 k (0.032 0.022)4.7 k
517 103 1.6 k 47 313.9
8.10 Complete Hybrid Equivalent Model
379
versus 323.125 using Av hfeRL/hie.
Io
hfe
110
(c) Eq. (8.107): Ai Ii
1 hoe RL
1 (20 S)(4.7 k)
110
100.55
1 0.094
versus 110 using simply hfe. Since 470 k
Zi, I i Ii and A i 100.55 also.
Vo
1
Zo hoe [hfehre/(hie Rs)]
Io
1
20 S [(110)(2 104)/(1.6 k 1 k)]
1
20 S 8.46 S
1
11.54 S
86.66 k
(d) Eq. (8.110):
which is greater than the value determined from 1/hoe 50 k.
Z o RCZo 4.7 k86.66 k 4.46 k
versus 4.7 k using only RC.
Note from the results above that the approximate solutions for Av and Zi were very
close to those calculated with the complete equivalent model. In fact, even Ai was off
by less than 10%. The higher value of Zo only contributed to our earlier conclusion
that Zo is often so high that it can be ignored compared to the applied load. However,
keep in mind that when there is a need to determine the impact of hre and hoe, the
complete hybrid equivalent model must be used, as described above.
The specification sheet for a particular transistor typically provides the commonemitter parameters as noted in Fig. 7.28. The next example will employ the same transistor parameters appearing in Fig. 8.49 in a pnp common-base configuration to introduce the parameter conversion procedure and emphasize the fact that the hybrid
equivalent model maintains the same layout.
EXAMPLE 8.14
For the common-base amplifier of Fig. 8.52, determine the following parameters using the complete hybrid equivalent model and compare the results to those obtained
using the approximate model.
(a) Zi and Z i.
(b) Ai and A i.
hie = 1.6 kΩ
hre = 2 × 10− 4
+
Rs
Ii'
1 kΩ
+
+
Ii
Z'i
2.2 kΩ
Zi
6V
–
Figure 8.52 Example 8.14.
380
Chapter 8
Io
3 kΩ
Vi
Vs
–
hfe = 110
hoe = 20 µS
BJT Small-Signal Analysis
Z'o
Zo
Vo
12 V
–
(c) Av.
(d) Zo and Z o.
Solution
The common-base hybrid parameters are derived from the common-emitter parameters using the approximate equations of Appendix A:
hie
1.6 k
hib 14.41 1 hfe
1 110
Note how closely the magnitude compares with the value determined from
hie
1.6 k
hib re 14.55 110
hiehoe
(1.6 k)(20 S)
hrb hre 2 104
1 hfe
1 110
0.883 104
hfe
110
hfb 0.991
1 hfe
1 110
hoe
20 S
hob 0.18 S
1 hfe
1 110
Substituting the common-base hybrid equivalent circuit into the network of Fig.
Rs
1 kΩ
Ii'
Ii
Z'i
Zi
+
+
3 kΩ
Vi
Vs
–
hib
e
Ie
c
14.41 Ω
Zo
+
0.883 × 10−4 Vo
hrb Vo
− 0.991Ie
hfb Ie
hob = 0.18 µ S
Z'o
2.2 kΩ
+
Vo
–
–
Thévenin
Io
–
b
b
Figure 8.53 Small-signal equivalent for the network of Fig. 8.52.
8.52 will then result in the small-signal equivalent network of Fig. 8.53. The Thévenin
network for the input circuit will result in RTh 3 k1 k 0.75 k for Rs in the
equation for Zo.
Vi
h f b h rb RL
(a) Eq. (8.109): Zi hib Ii
1 hob RL
(0.991)(0.883 104)(2.2 k)
14.41 1 (0.18 S)(2.2 k)
14.41 0.19 14.60 versus 14.41 using Zi hib.
Z i 3 kZi Zi 14.60 (b) Eq. (8.107):
hfb
Io
Ai 1 hob RL
Ii
8.10 Complete Hybrid Equivalent Model
381
0.991
1 (0.18 S)(2.2 k)
Since 3 k
0.991 hfb
Zi, I i Ii and A i Io/I i 1 also.
Vo
h fb RL
(c) Eq. (8.108): Av hib (hib hob hfbhrb)RL
Vi
(0.991)(2.2 k)
14.41 [(14.41 )(0.18 S) (0.991)(0.883 104)]2.2 k
149.25
versus 151.3 using Av hfb RL/hib.
(d) Eq. (8.110):
1
Zo hob[hf b hrb /(hib Rs)]
1
0.18 S [(0.991)(0.883 104)/(14.41 0.75 k)]
1
0.295 S
3.39 M
versus 5.56 M using Zo 1/hob. For Z o as defined by Fig. 8.53:
Z o RCZo 2.2 k3.39 M 2.199 k
versus 2.2 k using Z o RC.
8.11 SUMMARY TABLE
Now that the most familiar configurations of the small-signal transistor amplifiers
have been introduced, Table 8.1 is presented to review the general characteristics of
each for immediate recall. It must be absolutely clear that the values listed are simply typical values to establish a basis for comparison. The levels obtained in an actual analysis will most likely be different, and certainly different from one configuration to another. Being able to repeat most of the information in the table is an
important first step in developing a general familiarity with the subject matter. For instance, one should now be able to state with some assurance that the emitter-follower
configuration typically has a high input impedance, low output impedance, and a voltage gain slightly less than 1. There should be no need to perform a variety of calculations to recall salient facts such as those above. For the future, it will permit the
study of a network or system without becoming mathematically involved. The function of each component of a design will become increasingly familiar as general facts
such as those above become part of your background.
One obvious advantage of being able to recall general facts like the above is an
ability to check the results of a mathematical analysis. If the input impedance of a
common-base configuration is in the kilohm range, there is good reason to recheck
the analysis. However, on the other side of the coin, a result of 22 suggests that
the analysis may be correct.
382
Chapter 8
BJT Small-Signal Analysis
TABLE 8.1 Relative Levels for the Important Parameters of the CE, CB, and CC Transistor Amplifiers
Configuration
Zi
Fixed-bias:
VCC
RB
RC
VCC
Voltage-divider
bias:
RC
R1
Zo
Av
Ai
Medium (1 k)
Medium (2 k)
High (200)
RBre
RCro
(RCro)
re
re
RC
(RB 10re)
(ro 10RC)
High (100)
RBro
(ro RC)(RB re)
RC
re
(ro 10RC)
(ro 10RC ,
RB 10re)
High (50)
Medium (1 k)
Medium (2 k)
High (200)
R1R2re
RCro
RCro
re
(R1R2)ro
(ro RC)(R1R2 re)
RC
(ro 10RC)
R2
RE
Unbypassed
emitter bias:
CE
VCC
RC
RB
(ro 10RC)
Low (5)
High (50)
RBZb
RC
Zb (re RE)
(any level
of ro)
RC
re RE
RB
RB Zb
(RE
VCC
RB
RC
RE
re)
Low (20 )
Low ( 1)
High (50)
RBZb
REre
RE
RE re
RB
RB Zb
RBRE
(RE
Commonbase:
RE
re
(RE
re)
Low (20 )
Medium (2 k)
High (200)
Low (1)
REre
RC
RC
re
1
Medium (2 k)
High (200)
High (50)
RCRF
RC
re
RF
RF RC
re
VCC
(RE
Collector
feedback:
VCC
RF
1
re)
RC
VEE
re)
High (100 k)
Zb (re RE)
RE
(ro 10RC)
Medium (2 k)
(RE
Emitterfollower:
(R R2)
1
R1R2 re
High (100 k)
RBRE
RE
RC
re
re)
Medium (1 k)
RC
re
1
RC
RE
(ro 10RC)
(ro 10RC)
(ro 10RC)
RF
RC)
8.12 Troubleshooting
RF
RC
383
8.12 TROUBLESHOOTING
Although the terminology troubleshooting suggests that the procedures to be described
are designed simply to isolate a malfunction, it is important to realize that the same
techniques can be applied to ensure that a system is operating properly. In any case,
the testing, checking, or isolating procedures require an understanding of what to expect at various points in the network in both the dc and ac domains. In most cases, a
network operating correctly in the dc mode will also behave properly in the ac domain. In addition, a network providing the expected ac response is most likely biased
as planned. In a typical laboratory setting, both the dc and ac supplies are applied and
the ac response at various points in the network is checked with an oscilloscope as
shown in Fig. 8.54. Note that the black (gnd) lead of the oscilloscope is connected
directly to ground and the red lead is moved from point to point in the network, providing the patterns appearing in Fig. 8.54. The vertical channels are set in the ac mode
to remove any dc component associated with the voltage at a particular point. The
small ac signal applied to the base is amplified to the level appearing from collector
to ground. Note the difference in vertical scales for the two voltages. There is no ac
VCC
υ o (V)
RC
RB
C2
υo
0
C1
υi
+
υe
υ i (mV)
Rs
υe
0
t
0
t
Oscilloscope
RE
Vs
t
CE
–
(AC-GND-DC switch on AC)
Ground strap
Figure 8.54 Using the oscilloscope to measure and display various voltages of a BJT amplifier.
response at the emitter terminal due to the short-circuit characteristics of the capacitor at the applied frequency. The fact that vo is measured in volts and vi in millivolts
suggests a sizable gain for the amplifier. In general, the network appears to be operating properly. If desired, the dc mode of the multimeter could be used to check VBE
and the levels of VB, VCE, and VE to review whether they lie in the expected range.
Of course, the oscilloscope can also be used to compare dc levels simply by switching to the dc mode for each channel.
Needless to say, a poor ac response can be due to a variety of reasons. In fact,
there may be more than one problem area in the same system. Fortunately, however,
with time and experience, the probability of malfunctions in some areas can be predicted and an experienced person can isolate problem areas fairly quickly.
In general, there is nothing mysterious about the general troubleshooting process.
If you decide to follow the ac response, it is good procedure to start with the applied
384
Chapter 8
BJT Small-Signal Analysis
Figure 8.55 The waveforms resulting from a malfunction in the emitter area.
signal and progress through the system toward the load, checking critical points along
the way. An unexpected response at some point suggests that the network is fine up
to that area, thereby defining the region that must be investigated further. The waveform obtained on the oscilloscope will certainly help in defining the possible problems with the system.
If the response for the network of Fig. 8.54 is as appears in Fig. 8.55, the network
has a malfunction that is probably in the emitter area. An ac response across the emitter is unexpected, and the gain of the system as revealed by vo is much lower. Recall
for this configuration that the gain is much greater if RE is bypassed. The response
obtained suggests that RE is not bypassed by the capacitor and the terminal connections of the capacitor and the capacitor itself should be checked. In this case, a checking of the dc levels will probably not isolate the problem area since the capacitor has
an “open-circuit” equivalent for dc. In general, a prior knowledge of what to expect,
a familiarity with the instrumentation, and most important, experience are all factors
that contribute to the development of an effective approach to the art of troubleshooting.
8.13 PSPICE WINDOWS
Voltage-Divider Configuration Using the Software Transistor
Parameters
Now that the basic maneuvers for developing the network on the schematics grid have
been introduced, the current description will concentrate on the variations introduced
by the ac analysis.
Using schematics, the network of Fig. 8.9 (Example 8.2) is developed as shown
in Fig. 8.56. Note the ac source of 1 mV and the printer symbol at the output terminal of the network.
The sinusoidal ac source is listed in the SOURCE.slb library as VSIN. Once
8.13 PSpice Windows
385
Figure 8.56 Using PSpice Windows to analyze the network of
Figure 8.9 (Example 8.2).
placed on the diagram, double-clicking the symbol will result in the PartName: VSIN
dialog box with a list of options. Each choice can be made by double-clicking the desired quantity, which will then appear in the Name and Value rectangles at the top
of the box. The cursor appears in the Value box, and the desired value can be entered. After each entry, be sure to Save Attr to save the entered attribute. If done properly, the assigned value will appear in the listing.
For our analysis, the following choices will be made:
VAMPL 1mV (the peak value of the sinusoidal signal).
FREQ 10kHz (the frequency of interest).
PHASE 0 (no initial phase angle for the sinusoidal signal).
VOFF 0 (no dc offset voltage for the sinusoidal signal).
AC 1mV.
If you want to display the value of the ac signal, simply click on Change Display after saving the attribute. For instance, if AC 1mV was just saved and Change
Display was chosen, a Change Attribute dialog box would appear. Since AC is the
name and 1mV the value, choose Value Only, and only the 1mV will be displayed
after the sequence OK-OK.
The printer symbol on the collector of the transistor is listed as VPRINT1 under
the SPECIAL.slb library. When placed on the schematic, it dictates that the ac voltage at that point will be printed in the output file (*.out). Double-clicking on the
printer symbol will result in a PRINT1 dialog box, within which the following choices
should be made:
AC ok.
MAG ok.
PHASE ok.
After each entry, be sure to Save Attr or the computer will remind you. The above
choices can be listed next to the printer symbol on the schematic by simply clicking
the Change Display option and choosing the Display Value and Name for each item.
The transistor is obtained through the sequence Get New Part icon-LibrariesEVAL.slb-Q2N2222-OK-Place & Close. Since we will want the parameters of the
transistor to match those of the example as closely as possible, one must first click
on the transistor to put it in the active mode (red) and then choose Edit-Model-Edit
Instance Model (Text). Next, the beta (Bf) is set to 90 and Is is set to 2E-15A to result in a base-to-emitter voltage close to 0.7 V. This value of I is the result of nu386
Chapter 8
BJT Small-Signal Analysis
merous runs of the network to find that value of IS that provided a level of VBE closest to 0.7 V. For the remainder of this text, however, this chosen level of IS will remain the same. In most cases, it provides the desired results.
VIEWPOINTs have been inserted to display the three dc voltages of interest.
Since it has been used recently, VIEWPOINT can be found in the scroll listing at
the top right of the menu bar rather than by returning to the library listing.
Choosing the Setup Analysis icon will result in the Analysis Setup dialog box,
in which the AC Sweep must be chosen because of the applied ac source. Clicking
on AC Sweep will result in an AC Sweep and Noise Analysis dialog box, in which
Linear is chosen along with Total Pts: 1, Start Freq: 10kHz, and End Freq: 10kHz.
The result will be an analysis at only one frequency. Our initial interest will simply
be in the magnitude of the quantities of interest and not their shape or appearance.
Therefore, we should turn to Analysis-Probe Setup and choose Do not auto-run
Probe to save time getting to the desired results.
Clicking the Analysis icon will result in a PSpiceAD dialog box that will indicate the AC Analysis is finished. Note also the listing of the frequency applied at the
bottom of the dialog box. Within this box, if we choose File followed by Examine
Output, we will obtain a lengthy listing of input and output data on the analyzed network. Specific headings are duplicated in Fig. 8.57. Under Schematics Netlist, the
nodes assigned to the network are revealed. Note that ground is always defined as the
0 node and the assumed node of higher potential listed first. The transistor is listed
in the order Collector-Base-Emitter. Under BJT MODEL PARAMETERS, the defining parameters of the device are listed with the set values of IS 2E-15A and 90. Under SMALL-SIGNAL BIAS SOLUTION, the dc levels at the various nodes
are revealed, which compare directly with the VIEWPOINT values. In particular,
note that VBE is exactly 0.7 V.
The next listing, OPERATING POINT INFORMATION, reveals that even
though beta of the BJT MODEL PARAMETERS listing was set at 90, the operating conditions of the network resulted in a dc beta of 48.3 and an ac beta of 55. Fortunately, however, the voltage-divider configuration is less sensitive to changes in beta
in the dc mode, and the dc results are excellent. However, the drop in ac beta had an
effect on the resulting level of Vo: 296.1 mV versus the hand-written solution (with
ro 50 k) of 324.3 mV—a 9% difference. The results are certainly close, but probably not as close as one would like. A closer result (within 7%) could be obtained by
setting all the parameters of the device except IS and beta to zero. However, for the
moment, the impact of the remaining parameters has been demonstrated, and the results will be accepted as sufficiently close to the hand-written levels. Later in this
chapter, an ac model for the transistor will be introduced with results that will be an
exact match with the hand-written solution. The phase angle is 178° versus the ideal
of 180°—a very close match.
A plot of the output waveform can be obtained using the Probe option. The
sequence Analysis-Probe Setup-Automatically run Probe after simulation-OK
will result in a MicroSim Probe screen when the Analysis icon is chosen. However, if we follow this procedure without setting the horizontal scale, we will simply end up with a plot point of 296 mV at a frequency of 10 kHz. The horizontal
scale is set by the sequence Analysis-Setup-Transient with the AC Sweep disabled. Clicking the Transient option will result in a Transient dialog box, in which
a number of choices have to be made based on the waveform to be viewed. The
period of the applied signal of 10 kHz is 0.1 ms 100 s. The Print Step option
refers to the time interval between printing or plotting the results of the transient
analysis. For our example, we will choose 1 s to provide 100 plot points per cycle. The Final Time is the last instant the network’s response will be determined.
Our choice is 500 s or 0.5 ms to provide five full cycles of the waveform. The
No-Print Delay was chosen as 0 since all the capacitors are essentially short cir8.13 PSpice Windows
387
Figure 8.57 Output file for
the network of Figure 8.56.
388
cuits at 10 kHz. If we felt there was a transient phase between energizing the network and reaching a steady-state response, the No-Print Delay could be used to
effectively eliminate this period of time. The last choice of Step Ceiling sets a
maximum time period between response calculations for the system, which we will
set at 1 s. The time between calculations will be adjusted internally by the software package to ensure sufficient data at times when the response may change
faster than usual. However, they will never be separated by a time period greater
than that set by the Step Ceiling.
After Simulation, a MicroSim Probe screen will appear showing only the horizontal scale from 0 to 500 s as specified in the Transient dialog box. To obtain a
waveform, one can either choose Trace on the menu bar or the Trace icon (red pattern on a black axis). If the Trace on the menu bar is chosen, one must follow with
Add, and the Add Traces dialog box will appear. Using the icon results in the dialog box immediately. Now one must choose the waveform to be displayed from the
list of Simulation Output Variables. Since we want the collector-to-emitter voltage
of the transistor, V(Q1:c)—an option under Alias names—will be chosen, followed
by an OK. The result is the waveform of Fig. 8.58, with the waveform riding on the
Figure 8.58 Voltage vc for the network of Figure 8.56.
dc level of 13.45 V. The range of the vertical axis was automatically chosen by the
computer. Five full cycles of the output waveform are displayed (as we expected) with
100 data points for each cycle. If you would like to see the data points (as shown in
Fig. 8.58), simply turn to Tools-Options-Probe Options and choose Mark Data
Points. Click OK, and the data points will appear. Using the scale of the graph, the
peak-to-peak value of the curve is approximately 13.76 V 13.16 V 0.6 V 600 mV, resulting in a peak value of 300 mV. Since a 1-mV signal was applied, the
gain is 300, or very close to the values displayed above.
If a comparison is to be made between the input and output voltages on the same
graph, the Add Y-Axis option under Plot can be chosen. After it is triggered, choose
8.13 PSpice Windows
389
Figure 8.59 The voltages vc and vs for the network of Figure 8.56.
the Add Trace icon and select V(Vs:). The result is that both waveforms will appear on the same screen, each with their own vertical scale. Labels can be added to
the waveforms as shown in Fig. 8.59 using Tools-Label-Text. A Text Label dialog
box will appear, in which the desired text can be entered. Click OK, and it can be
placed with the mouse in any location on the graph. Lines can also be added with
Tools-Label-Line. A pencil will appear, which can be used to draw the line with a
left-click at the starting point and another click when the line is in place. Each plot
Figure 8.60 Two separate plots of vc and vs in Figure 8.56.
390
Chapter 8
BJT Small-Signal Analysis
can be printed with File-Print-Copies-OK.
If two separate graphs are preferred, we can choose the Plot option and select
Add Plot after V has been displayed. Upon selection, another graph will appear, waiting for the next choice. The sequence Trace-Add-V(Vs:) will then result in the graphs
of Fig. 8.60. The labels Vs and Vc were added using the Tools option. If further operations are to be performed on either graph, the SEL defines the active plot.
The last waveform of this section will demonstrate the use of the Cursor option
that can be called up using the Tools menu choice or the Cursor Point icon (having
a graph with an arrow drawn from the graph to the vertical axis). The sequence ToolsCursor-Display will result in a line at the dc level of 13.453 V, as shown in the dialog box at the bottom right of the graph of Fig. 8.61. Left-clicking on the mouse
once will result in a horizontal and vertical line intersecting at some point on the
curve. By clicking on the vertical line and holding it down, the vertical line and corresponding horizontal line (on the graph) can be moved across the waveform. At each
point, the vertical and horizontal intersections will appear in the dialog box. If moved
to the first peak value, A1 will be at 13.754 V and 74.825 s. By right-clicking
on the mouse, a second intersection, defined by A2, appears, which also has its location registered in the dialog box. These intersecting lines are moved by holding down
Figure 8.61 Demonstrating the use of cursors to read specific points on a plot.
the right side of the mouse. The remaining information on the third line of the box
is the difference between the two intersections on the two axis. If A2 is set at the
bottom of the waveform as shown in Fig. 8.61, it will read 13.162 V at 125.17 s,
resulting in a difference between the two of 591.999 mV, or 0.592 V vertically and
50.35 s horizontally. This is as expected, because the peak-to-peak value matches
the 2 0.296 V 0.592 V obtained earlier. The time interval is essentially 1/4 of
the total period (200 s) of the waveform. The labels A1 and A2 were added using
the Tools-Label-Text sequence or the ABC text icon.
The peak and minimum values for the graph of Fig. 8.61 can also be found using the icons appearing in the top right region of the menu bar. Once the desired wave8.13 PSpice Windows
391
form is obtained and the sequence Tools-Cursor-Display applied or the Toggle cursor icon (the icon in the center region of the menu bar with the black dashed axis and
red curve passing through the origin) is chosen, the six icons to the right of the Toggle cursor icon will change to a color pattern indicating they are ready for use. Clicking on the icon with the intersection at the top will automatically place the A1 intersection at the top of the curve. Clicking the next icon to the right will place the
intersection at the bottom (trough) of the curve. The next icon will place the intersec-
B
C
B
C
βre
Ib
βre
βIb
GAIN = β
F1
B
βre
βIb
Ib
C
βIb
Ib
F
E
E
E
Figure 8.62 Using a controlled source to represent the transistor of Figure 8.56.
tion at the steepest slope and the next at the minimum value (matching the trough value).
Voltage-Divider Configuration—Controlled Source Substitution
The results obtained for any analysis using the transistors provided in the software
package will always be different from those obtained with an equivalent model that
only includes the effect of beta and re. This was demonstrated for the network of Fig.
8.56. If a solution is desired that is limited to the approximate model, then the transistor must be represented by a model such as appearing in Fig. 8.62.
For Example 8.2, is 90, with re 1.66 k. The current controlled current
source (CCCS) is found in the ANALOG.slb library as Part F. When you click on
F, the Description above will read Current-controlled current source. After OKPlace & Close, the graphical symbol for the CCCS will appear on the screen as
shown in Fig. 8.63. Since re does not appear within the basic structure of the CCCS,
it must be added in series with the controlling current indicated by the arrow on the
Figure 8.63 Substituting the controlled source of Figure 8.62 for the transistor
of Figure 8.56.
392
Chapter 8
BJT Small-Signal Analysis
§ 8.2 Common-Emitter Fixed-Bias Configuration
1. For
(a)
(b)
(c)
(d)
PROBLEMS
the network of Fig. 8.64:
Determine Zi and Zo.
Find Av and Ai.
Repeat part (a) with ro 20 k.
Repeat part (b) with ro 20 k.
12 V
220 kΩ
2.2 kΩ
Io
Vo
Vi
Zo
β = 60
ro = 40 kΩ
Ii
Zi
Figure 8.64 Problems 1 and 21
2. For the network of Fig. 8.65, determine VCC for a voltage gain of Av 200.
VCC
4.7 kΩ
1 MΩ
Vo
β = 90
ro = ∞ Ω
Vi
Figure 8.65 Problem 2
* 3. For
(a)
(b)
(c)
(d)
the network of Fig. 8.66:
Calculate IB, IC, and re.
Determine Zi and Zo.
Calculate Av and Ai.
Determine the effect of ro 30 k on Av and Ai.
10 V
4.3 kΩ
Io
Vo
Zo
Vi
Ii
390 kΩ
β = 100
ro = 60 kΩ
Zi
+10 V
Figure 8.66 Problem 3
Problems
393
§ 8.3 Voltage-Divider Bias
4. For
(a)
(b)
(c)
(d)
the network of Fig. 8.67:
Determine re.
Calculate Zi and Zo.
Find Av and Ai.
Repeat parts (b) and (c) with ro 25 k.
VCC = 16 V
VCC
39 kΩ
3.9 kΩ
Io
1 µF
1 µF
Vi
3.3 kΩ
82 kΩ
Vo
Zo
β = 100
ro = 50 kΩ
Ii
Vo
CC
CC
4.7 kΩ
Zi
β = 100
ro = ∞ Ω
Vi
5.6 kΩ
1.2 kΩ
10 µF
Figure 8.67 Problem 4
CE
1 kΩ
Figure 8.68 Problem 5
5. Determine VCC for the network of Fig. 8.68 if Av 160 and ro 100 k.
6. For
(a)
(b)
(c)
the network of Fig. 8.69:
Determine re.
Calculate VB and VC.
Determine Zi and Av Vo /Vi.
VCC = 20 V
6.8 kΩ
220 kΩ
VB
Vi
CC
Zi
VC
Vo
CC
β = 180
ro = 50 kΩ
20 V
56 kΩ
2.2 kΩ
CE
390 kΩ
2.2 kΩ
Io
Vo
Figure 8.69 Problem 6
β = 140
ro = 100 kΩ
Vi
§ 8.4 CE Emitter-Bias Configuration
7. For
(a)
(b)
(c)
(d)
394
the network of Fig. 8.70:
Determine re.
Find Zi and Zo.
Calculate Av and Ai.
Repeat parts (b) and (c) with ro 20 k.
Chapter 8
BJT Small-Signal Analysis
Ii
Zi
1.2 kΩ
Figure 8.70 Problems 7 and 9
Zo
8. For the network of Fig. 8.71, determine RE and RB if Av 10 and re 3.8 . Assume that
Zb RE.
9. Repeat Problem 7 with RE bypassed. Compare results.
* 10. For
(a)
(b)
(c)
the network of Fig. 8.72:
Determine re.
Find Zi and Av.
Calculate Ai.
22 V
5.6 kΩ
20 V
Io
330 kΩ
Vo
Ii
CC
β = 80
ro = 40 kΩ
Vi
8.2 kΩ
CC
RB
Vo
Zi
1.2 kΩ
β = 120
ro = 80 kΩ
Vi
0.47 kΩ
RE
CE
Figure 8.72 Problem 10
Figure 8.71 Problem 8
§ 8.5 Emitter-Follower Configuration
11. For
(a)
(b)
(c)
* 12. For
(a)
(b)
(c)
the network of Fig. 8.73:
Determine re and re.
Find Zi and Zo.
Calculate Av and Ai.
the network of Fig. 8.74:
Determine Zi and Zo.
Find Av.
Calculate Vo if Vi 1 mV.
16 V
12 V
Ii
270 kΩ
β = 110
ro = 50 kΩ
Vi
β = 120
ro = 40 kΩ
Vi
Ii
Zi
Io
2.7 kΩ
Vo
Zi
Vo
390 kΩ
Io
5.6 kΩ
Zo
Zo
−8 V
Figure 8.73 Problem 11
Figure 8.74 Problem 12
Problems
395
* 13. For
(a)
(b)
(c)
(d)
the network of Fig. 8.75:
Calculate IB and IC.
Determine re.
Determine Zi and Zo.
Find Av and Ai.
VCC = 20 V
56 kΩ
β = 200
ro = 40 kΩ
Vi
Ii
Vo
Io
8.2 kΩ
2 kΩ
Figure 8.75 Problem 13
§ 8.6 Common-Base Configuration
14. For
(a)
(b)
(c)
the common-base configuration of Fig. 8.76:
Determine re.
Find Zi and Zo.
Calculate Av and Ai.
+6 V
−10 V
6.8 kΩ
4.7 kΩ
Ii
Io
Vi
Zi
Vo
α = 0.998
ro = 1 MΩ
Zo
Figure 8.76 Problem 14
* 15. For the network of Fig. 8.77, determine Av and Ai.
8V
3.6 kΩ
Io
Vo
β = 75
ro = ∞ Ω
Vi
Ii
3.9 kΩ
−5 V
396
Chapter 8
BJT Small-Signal Analysis
Figure 8.77 Problem 15
§ 8.7 Collector Feedback Configuration
16. For
(a)
(b)
(c)
12 V
the collector FB configuration of Fig. 8.78:
Determine re.
Find Zi and Zo.
Calculate Av and Ai.
Io
3.9 kΩ
220 kΩ
Vo
Zo
Vi
β = 120
ro = 40 kΩ
Ii
Zi
Figure 8.78 Problem 16
* 17. Given re 10 , 200, Av 160, and Ai 19 for the network of Fig. 8.79, determine
RC, RF, and VCC.
VCC
RC
RF
Vo
re = 10 Ω
Vi
β = 200
ro = 80 kΩ
Figure 8.79 Problem 17
* 18. For
(a)
(b)
(c)
(d)
the network of Fig. 8.30:
Derive the approximate equation for Av.
Derive the approximate equation for Ai.
Derive the approximate equations for Zi and Zo.
Given RC 2.2 k, RF 120 k, RE 1.2 k, 90, and VCC 10 V, calculate the
magnitudes of Av, Ai, Zi, and Zo using the equations of parts (a) through (c).
§ 8.8 Collector DC Feedback Configuration
19. For the network of Fig. 8.80:
(a) Determine Zi and Zo.
(b) Find Av and Ai.
9V
Io
1.8 kΩ
39 kΩ
22 kΩ
10 µ F
Ii
Vi
1 µF
1 µF
Vo
Zo
β = 80
ro = 45 kΩ
Zi
Figure 8.80 Problem 19
Problems
397
§ 8.9 Approximate Hybrid Equivalent Circuit
20. (a) Given 120, re 4.5 , and ro 40 k, sketch the approximate hybrid equivalent
circuit.
(b) Given hie 1 k, hre 2104, hfe 90, and hoe 20 S, sketch the re model.
21. For
(a)
(b)
(c)
(d)
(e)
(f)
(g)
the network of Problem 1:
Determine re.
Find hfe and hie.
Find Zi and Zo using the hybrid parameters.
Calculate Av and Ai using the hybrid parameters.
Determine Zi and Zo if hoe 50 S.
Determine Av and Ai if hoe 50 S.
Compare the solutions above with those of Problem 1. (Note: The solutions are available
in Appendix E if Problem 1 was not performed.)
22. For
(a)
(b)
(c)
the network of Fig. 8.81:
Determine Zi and Zo.
Calculate Av and Ai.
Determine re and compare re to hie.
18 V
2.2 kΩ
68 kΩ
Io
Ii
5 µF
Zo
5 µF
12 kΩ
Zi
1.2 kΩ
Vo
hfe = 180
hie = 2.75 kΩ
hoe = 25 µS
10 µF
Figure 8.81 Problems 22 and 24
* 23. For
(a)
(b)
(c)
the common-base network of Fig. 8.82:
Determine Zi and Zo.
Calculate Av and Ai.
Determine , , re, and ro.
hfb = −0.992
hib = 9.45 Ω
hob = 1 µ A/V
Ii
+
Io
10 µ F
1.2 kΩ
2.7 kΩ
4V
12 V
10 µF
Vi
Vo
Zi
–
Chapter 8
Zo
–
Figure 8.82 Problem 23
398
+
BJT Small-Signal Analysis
§ 8.10 Complete Hybrid Equivalent Model
4
* 24. Repeat parts (a) and (b) of Problem 22 with hre 2 10 and compare results.
* 25. For the network of Fig. 8.83, determine:
(a) Zi.
(b) Av.
(c) Ai Io/Ii.
(d) Zo.
20 V
2.2 kΩ
470 kΩ
5 µF
Vo
5 µF
Ii
1 kΩ
+
Io
+
Zo
Vi
Vs
1.2 kΩ
–
Zi
hfe = 140
hie = 0.86 kΩ
hre = 1.5 × 10− 4
hoe = 25 µS
10 µ F
–
Figure 8.83 Problem 25
* 26. For
(a)
(b)
(c)
(d)
the common-base amplifier of Fig. 8.84, determine:
Zi.
Ai.
Av.
Zo.
hib = 9.45 Ω
hfb = −0.997
hob = 0.5 µ A/V
hrb = 1 × 10− 4
Ii
0.6 kΩ
5 µF
Io
+
+
5 µF
+
1.2 kΩ
Vs
Zi
2.2 kΩ
Vi
–
Zo
4V
–
Vo
14 V
–
Figure 8.84 Problem 26
§ 8.12 Troubleshooting
* 27. Given the network of Fig. 8.85:
(a) Determine if the system is operating properly based on the voltage-divider bias levels and
expected waveforms for vo and vE.
(b) Determine the reason for the dc levels obtained and why the waveform for vo was obtained.
Problems
399
VCC = 14 V
υi (mV)
RC
R1
υo (V)
2.2 kΩ
150 kΩ
10 µ F
t
0
10 µ F
vo
VB = 6.22 V
β = 70
+
C1
–
R2
+
υe
VBE = 0.7 V
Rs
t
0
C2
0
t
39 kΩ
RE
1.5 kΩ
10 µ F
Vs
–
Figure 8.85 Problem 27
§ 8.13 PSpice Windows
28. Using PSpice Windows, determine the gain for the network of Fig. 8.6. Use Probe to display
the input and output waveforms.
29. Using PSpice Windows, determine the gain for the network of Fig. 8.13. Use Probe to display
the input and output waveforms.
30. Using PSpice Windows, determine the gain for the network of Fig. 8.25. Use Probe to display
the input and output waveforms.
*Please Note: Asterisks indicate more difficult problems.
400
Chapter 8
BJT Small-Signal Analysis
gm
CHAPTER
FET Small-Signal
Analysis
9
9.1 INTRODUCTION
Field-effect transistor amplifiers provide an excellent voltage gain with the added feature of a high input impedance. They are also considered low-power consumption
configurations with good frequency range and minimal size and weight. Both JFET
and depletion MOSFET devices can be used to design amplifiers having similar voltage gains. The depletion MOSFET circuit, however, has a much higher input impedance than a similar JFET configuration.
While a BJT device controls a large output (collector) current by means of a relatively small input (base) current, the FET device controls an output (drain) current
by means of a small input (gate-voltage) voltage. In general, therefore, the BJT is a
current-controlled device and the FET is a voltage-controlled device. In both cases,
however, note that the output current is the controlled variable. Because of the high
input characteristic of FETs, the ac equivalent model is somewhat simpler than that
employed for BJTs. While the BJT had an amplification factor (beta), the FET has
a transconductance factor, gm.
The FET can be used as a linear amplifier or as a digital device in logic circuits.
In fact, the enhancement MOSFET is quite popular in digital circuitry, especially in
CMOS circuits that require very low power consumption. FET devices are also widely
used in high-frequency applications and in buffering (interfacing) applications. Table
9.1, located at the end of the chapter, provides a summary of FET small-signal amplifier circuits and related formulas.
While the common-source configuration is the most popular providing an inverted,
amplified signal, one also finds common-drain (source-follower) circuits providing
unity gain with no inversion and common-gate circuits providing gain with no inversion. As with BJT amplifiers, the important circuit features described in this chapter
include voltage gain, input impedance, and output impedance. Due to the very high
input impedance, the input current is generally assumed to be 0 A and the current
gain is an undefined quantity. While the voltage gain of an FET amplifier is generally less than that obtained using a BJT amplifier, the FET amplifier provides a much
higher input impedance than that of a BJT configuration. Output impedance values
are comparable for both BJT and FET circuits.
FET ac amplifier networks can also be analyzed using computer software. Using
PSpice, one can perform a dc analysis to obtain the circuit bias conditions and an ac
analysis to determine the small-signal voltage gain. Using PSpice transistor models,
401
gm
one can analyze the circuit using specific transistor models. On the other hand, one
can develop a program using a language such as BASIC that can perform both the dc
and ac analyses and provide the results in a very special format.
9.2 FET SMALL-SIGNAL MODEL
The ac analysis of an FET configuration requires that a small-signal ac model for the
FET be developed. A major component of the ac model will reflect the fact that an
ac voltage applied to the input gate-to-source terminals will control the level of current from drain to source.
The gate-to-source voltage controls the drain-to-source (channel) current of an
FET.
Recall from Chapter 6 that a dc gate-to-source voltage controlled the level
of dc drain current through a relationship known as Shockley’s equation: ID IDSS(1 VGS/VP)2. The change in collector current that will result from a change
in gate-to-source voltage can be determined using the transconductance factor gm
in the following manner:
ID gm VGS
(9.1)
The prefix trans- in the terminology applied to gm reveals that it establishes a relationship between an output and input quantity. The root word conductance was chosen because gm is determined by a voltage-to-current ratio similar to the ratio that defines the conductance of a resistor G 1/R I/V.
Solving for gm in Eq. (9.1), we have:
ID
gm VGS
(9.2)
Graphical Determination of gm
If we now examine the transfer characteristics of Fig. 9.1, we find that gm is actually
the slope of the characteristics at the point of operation. That is,
y
ID
gm m x
VGS
(9.3)
Following the curvature of the transfer characteristics, it is reasonably clear that
the slope and, therefore, gm increase as we progress from VP to IDSS. Or, in other
words, as VGS approaches 0 V, the magnitude of gm increases.
ID
IDSS
gm ≡
∆ID
(= Slope at Q-point)
∆VGS
Q-Point
∆ID
∆VGS
VP
402
Chapter 9
FET Small-Signal Analysis
0
VGS
Figure 9.1 Definition of gm
using transfer characteristic.
gm
Equation (9.2) reveals that gm can be determined at any Q-point on the transfer
characteristics by simply choosing a finite increment in VGS (or in ID) about the
Q-point and then finding the corresponding change in ID (or VGS, respectively). The
resulting changes in each quantity are then substituted in Eq. (9.2) to determine gm.
Determine the magnitude of gm for a JFET with IDSS 8 mA and VP 4 V at the
following dc bias points:
(a) VGS 0.5 V.
(b) VGS 1.5 V.
(c) VGS 2.5 V.
EXAMPLE 9.1
Solution
The transfer characteristics are generated as Fig. 9.2 using the procedure defined in
Chapter 6. Each operating point is then identified and a tangent line is drawn at each
point to best reflect the slope of the transfer curve in this region. An appropriate increment is then chosen for VGS to reflect a variation to either side of each Q-point.
Equation (9.2) is then applied to determine gm.
2.1 m A
ID
(a) gm 3.5 mS
0.6 V
VGS
1.8 m A
ID
(b) gm 2.57 mS
0.7 V
VGS
1.5 m A
ID
(c) gm 1.5 mS
1.0 V
VGS
Note the decrease in gm as VGS approaches VP.
ID (mA)
8
7
gm at − 0.5 V
(
ID = 8 mA 1 −
VGS
−4 V
)
2
0.6 V
4
1.8 mA
3
2
gm at −2.5 V
−3
2.1 mA
5
gm at −1.5 V
−4
VP
6
0.7 V
1.5 mA
−2
−1
1.0 V
1
0
VGS (V)
Figure 9.2 Calculating gm at
various bias points.
Mathematical Definition of gm
The graphical procedure just described is limited by the accuracy of the transfer plot
and the care with which the changes in each quantity can be determined. Naturally,
the larger the graph the better the accuracy, but this can then become a cumbersome
9.2 FET Small-Signal Model
403
gm
problem. An alternative approach to determining gm employs the approach used to
find the ac resistance of a diode in Chapter 1, where it was stated that:
The derivative of a function at a point is equal to the slope of the tangent line
drawn at that point.
If we therefore take the derivative of ID with respect to VGS (differential calculus) using Shockley’s equation, an equation for gm can be derived as follows:
ID
gm VGS
dID
dVGS
Q-pt.
V
d
IDSS 1 GS
VP
dVGS
V
d
IDSS 1 GS
VP
dVGS
Q-pt.
2
2
V
V
d
2IDSS 1 GS 1 GS
VP dVGS
VP
V
V
d
1 dVGS
1
2IDSS 1 GS (1) 2IDSS 1 GS 0 VP dVGS
VP
VP dVGS
VP
2I SS
V
gm D
1 GS
VP
VP
and
(9.4)
where VP denotes magnitude only to ensure a positive value for gm.
It was mentioned earlier that the slope of the transfer curve is a maximum at
VGS 0 V. Plugging in VGS 0 V into Eq. (9.4) will result in the following equation
for the maximum value of gm for a JFET in which IDSS and VP have been specified:
2I SS
0
1 gm D
VP
VP
2I SS
gm0 D
VP
and
(9.5)
where the added subscript 0 reminds us that it is the value of gm when VGS 0 V.
Equation (9.4) then becomes
V
gm gm0 1 GS
VP
EXAMPLE 9.2
(9.6)
For the JFET having the transfer characteristics of Example 9.1:
(a) Find the maximum value of gm.
(b) Find the value of gm at each operating point of Example 9.1 using Eq. (9.6) and
compare with the graphical results.
Solution
2I SS
2(8 mA)
(a) gm0 D
4 mS
VP
4V
(b) At VGS 0.5 V,
(maximum possible value of gm)
V
0.5 V
gm gm0 1 GS 4 mS 1 3.5 mS
VP
4 V
404
Chapter 9
FET Small-Signal Analysis
(versus 3.5 mS
graphically)
gm
At VGS 1.5 V,
V
1.5 V
gm gm0 1 GS 4 mS 1 2.5 mS
VP
4 V
(versus 2.57 mS
graphically)
At VGS 2.5 V,
V
2.5 V
gm gm0 1 GS 4 mS 1 1.5 mS
VP
4 V
(versus 1.5 mS
graphically)
The results of Example 9.2 are certainly sufficiently close to validate Eq. (9.4)
through (9.6) for future use when gm is required.
On specification sheets, gm is provided as yfs where y indicates it is part of an admittance equivalent circuit. The f signifies forward transfer parameter, and the s reveals that it is connected to the source terminal.
In equation form,
gm yfs
(9.7)
For the JFET of Fig. 5.18, yfs ranges from 1000 to 5000 S or 1 to 5 mS.
Plotting gm vs. VGS
V
Since the factor 1 GS of Eq. (9.6) is less than 1 for any value of VGS other than
VP
V
0 V, the magnitude of gm will decrease as VGS approaches VP and the ratio GS
VP
increases in magnitude. At VGS VP, gm gm0(1 1) 0. Equation (9.6) defines a
straight line with a minimum value of 0 and a maximum value of gm as shown by the
plot of Fig. 9.3.
gm (S)
gm0
gm0
2
VP
VP
2
0
VGS (V)
Figure 9.3 Plot of gm vs. VGS.
Figure 9.3 also reveals that when VGS is one-half the pinch-off value, gm will be
one-half the maximum value.
EXAMPLE 9.3
Plot gm vs. VGS for the JFET of examples 9.1 and 9.2.
Solution
Note Fig. 9.4.
9.2 FET Small-Signal Model
405
gm
gm (S)
4 mS
2 mS
−4 V
−2 V
0
VGS (V)
Figure 9.4 Plot of gm vs. VGS for
a JFET with IDSS 8 mA and
VP 4 V.
Impact of ID on gm
A mathematical relationship between gm and the dc bias current ID can be derived by
noting that Shockley’s equation can be written in the following form:
V
1 GS VP
I
D
IDSS
(9.8)
Substituting Eq. (9.8) into Eq. (9.6) will result in
V
gm gm0 1 GS gm0
VP
I
D
IDSS
Using Eq. (9.9) to determine gm for a few specific values of ID, the results are
(a) If ID IDSS,
gm gm0
I SS
D
gm0
IDSS
(b) If ID IDSS/2,
gm gm0
IDSS/2
0.707gm0
IDSS
(c) If ID IDSS/4,
gm gm0
EXAMPLE 9.4
Plot gm vs. ID for the JFET of Examples 9.1 through 9.3.
Solution
See Fig. 9.5.
406
IDSS/4
g
m0
0.5gm0
IDSS
2
Chapter 9
FET Small-Signal Analysis
(9.9)
gm
gm (S)
4
3
2
4 mS
2.83 mS
2 mS
1
0
1
2
IDSS
4
3
4
IDSS
5
6
7
8
IDSS
9
10
ID (mA)
2
Figure 9.5 Plot of gm vs. ID for a JFET with IDSS 8 mA and VGS 4 V.
The plots of Examples 9.3 and 9.4 clearly reveal that the highest values of gm are
obtained when VGS approaches 0 V and ID its maximum value of IDSS.
FET Input Impedance Zi
The input impedance of all commercially available FETs is sufficiently large to assume that the input terminals approximate an open circuit. In equation form,
Zi (FET) (9.10)
For a JFET a practical value of 109 (1000 M) is typical, while a value of 1012
to 1015 is typical for MOSFETs.
FET Output Impedance Zo
The output impedance of FETs is similar in magnitude to that of conventional BJTs.
On FET specification sheets, the output impedance will typically appear as yos with
the units of S. The parameter yos is a component of an admittance equivalent circuit, with the subscript o signifying an output network parameter and s the terminal
(source) to which it is attached in the model. For the JFET of Fig. 5.18, yos has a
range of 10 to 50 S or 20 k (R 1/G 1/50 S) to 100 k (R 1/G 1/10 S).
In equation form,
1
Zo (FET) rd yos
(9.11)
The output impedance is defined on the characteristics of Fig. 9.6 as the slope of
the horizontal characteristic curve at the point of operation. The more horizontal the
curve, the greater the output impedance. If perfectly horizontal, the ideal situation is
on hand with the output impedance being infinite (an open circuit)—an often applied
approximation.
In equation form,
VDS
rd ID
(9.12)
VGSconstant
9.2 FET Small-Signal Model
407
gm
ID (mA)
VGS = 0 V
rd =
∆VDS
∆ID
VGS = constant at −1 V
VGS
Q-point
−1 V
∆ ID
∆VDS
−2 V
0
VDS (V)
Figure 9.6 Definition of rd using FET drain characteristics.
Note the requirement when applying Eq. (9.12) that the voltage VGS remain constant
when rd is determined. This is accomplished by drawing a straight line approximating the VGS line at the point of operation. A VDS or ID is then chosen and the other
quantity measured off for use in the equation.
EXAMPLE 9.5
Determine the output impedance for the FET of Fig. 9.7 for VGS 0 V and VGS 2 V at VDS 8 V.
ID (mA)
VGS = 0 V
8
7
∆VDS = 5 V
6
∆ ID = 0.2 mA
VGS = −1 V
5
4
3
∆ VDS = 8 V
2
VGS = −2 V
∆ ID = 0.1 mA
VGS = −3 V
VGS = −4 V
1
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14
VDS (V)
Figure 9.7 Drain characteristics used to calculate rd in Example 9.5.
Solution
For VGS 0 V, a tangent line is drawn and VDS is chosen as 5 V, resulting in a ID
of 0.2 mA. Substituting into Eq. (9.12),
VDS
rd ID
5V
25 k
0.2 mA
VGS0 V
For VGS 2 V, a tangent line is drawn and VDS is chosen as 8 V, resulting in a
ID of 0.1 mA. Substituting into Eq. (9.12),
408
Chapter 9
FET Small-Signal Analysis
gm
VDS
rd ID
8V
80 k
0.1
mA
VGS2 V
revealing that rd does change from one operating region to another, with lower values typically occurring at lower levels of VGS (closer to 0 V).
FET AC Equivalent Circuit
Now that the important parameters of an ac equivalent circuit have been introduced
and discussed, a model for the FET transistor in the ac domain can be constructed.
The control of Id by Vgs is included as a current source gmVgs connected from drain
to source as shown in Fig. 9.8. The current source has its arrow pointing from drain
to source to establish a 180° phase shift between output and input voltages as will
occur in actual operation.
G
D
+
gmVgs
Vgs
rd
−
S
S
Figure 9.8 FET ac equivalent circuit.
The input impedance is represented by the open circuit at the input terminals and
the output impedance by the resistor rd from drain to source. Note that the gate to
source voltage is now represented by Vgs (lower-case subscripts) to distinguish it from
dc levels. In addition, take note of the fact that the source is common to both input
and output circuits while the gate and drain terminals are only in “touch” through the
controlled current source gmVgs.
In situations where rd is ignored (assumed sufficiently large to other elements of
the network to be approximated by an open circuit), the equivalent circuit is simply
a current source whose magnitude is controlled by the signal Vgs and parameter gm —
clearly a voltage-controlled device.
Given yfs 3.8 mS and yos 20 S, sketch the FET ac equivalent model.
EXAMPLE 9.6
Solution
gm yfs 3.8 mS
and
1
1
rd 50 k
yos
20 S
resulting in the ac equivalent model of Fig. 9.9.
G
D
+
3.8 × 10−3 Vgs
Vgs
S
50 kΩ
−
S
Figure 9.9 FET ac equivalent model for Example 9.6.
9.2 FET Small-Signal Model
409
gm
9.3 JFET FIXED-BIAS CONFIGURATION
Now that the FET equivalent circuit has been defined, a number of fundamental FET
small-signal configurations will be investigated. The approach will parallel the ac
analysis of BJT amplifiers with a determination of the important parameters of Zi, Zo,
and Av for each configuration.
The fixed-bias configuration of Fig. 9.10 includes the coupling capacitors C1 and
C2 that isolate the dc biasing arrangement from the applied signal and load; they act
as short-circuit equivalents for the ac analysis.
+VDD
RD
C2
D
C1
Vo
G
Vi
RG
S
Zo
–
Zi
VGG
+
Figure 9.10 JFET fixed-bias configuration.
Once the level of gm and rd are determined from the dc biasing arrangement, specification sheet, or characteristics, the ac equivalent model can be substituted between
the appropriate terminals as shown in Fig. 9.11. Note that both capacitors have the
short-circuit equivalent because the reactance XC 1/(2 fC) is sufficiently small compared to other impedance levels of the network, and the dc batteries VGG and VDD are
set to zero volts by a short-circuit equivalent.
XC ≈ 0 Ω
XC ≈ 0 Ω
1
2
G
Vi
D
RG
gmVgs
Zi
Battery VGG
replaced by
short
rd
Vo
RD
Zo
Battery VDD
replaced by
short
S
Figure 9.11 Substituting the JFET ac equivalent circuit unit into the network of Fig. 9.10.
The network of Fig. 9.11 is then carefully redrawn as shown in Fig. 9.12. Note
the defined polarity of Vgs, which defines the direction of gmVgs. If Vgs is negative,
the direction of the current source reverses. The applied signal is represented by Vi
and the output signal across RD by Vo.
Zi: Figure 9.12 clearly reveals that
Zi RG
because of the open-circuit equivalence at the input terminals of the JFET.
410
Chapter 9
FET Small-Signal Analysis
(9.13)
gm
D
G
+
Vi
+
Zi
RG
+
gmVgs
Vgs
–
–
rd
RD
Zo
Vo
–
S
Figure 9.12 Redrawn network of Fig. 9.11.
Zo: Setting Vi 0 V as required by the definition of Zo will establish Vgs as 0
V also. The result is gmVgs 0 mA, and the current source can be replaced by an
open-circuit equivalent as shown in Fig. 9.13. The output impedance is
Zo RD rd
(9.14)
If the resistance rd is sufficiently large (at least 101) compared to RD, the approximation rd RD RD can often be applied and
Zo RD
(9.15)
rd 10RD
D
gmVgs = 0 mA
rd
RD
Zo
S
Figure 9.13 Determining Zo.
Av: Solving for Vo in Fig. 9.12, we find
Vo gmVgs(rd RD)
Vgs Vi
but
Vo gmVi(rd RD)
and
so that
Vo
gm(rd RD)
Av Vi
If rd
(9.16)
10RD:
Vo
gmRD
Av Vi
(9.17)
rd 10RD
Phase Relationship: The negative sign in the resulting equation for Av clearly reveals a phase shift of 180° between input and output voltages.
9.3 JFET Fixed-Bias Configuration
411
gm
EXAMPLE 9.7
The fixed-bias configuration of Example 6.1 had an operating point defined by VGSQ
2 V and IDQ 5.625 mA, with IDSS 10 mA and VP 8 V. The network is
redrawn as Fig. 9.14 with an applied signal Vi. The value of yos is provided as 40 S.
(a) Determine gm.
(b) Find rd.
(c) Determine Zi.
(d) Calculate Zo.
(e) Determine the voltage gain Av.
(f) Determine Av ignoring the effects of rd.
20 V
2 kΩ
D
+
C1
IDSS = 10 mA
VP = −8 V
G
+
1 MΩ
Vi
C2
S
Zo
Zi
Vo
2V
–
–
Figure 9.14 JFET configuration for Example 9.7.
Solution
2I SS
2(10 mA)
(a) gm0 D
2.5 mS
VP
8V
VGSQ
(2 V)
gm gm0 1 2.5 mS 1 1.88 mS
(8 V)
VP
1
1
(b) rd 25 k
yos
40 S
(c) Zi RG 1 M
(d) Zo RD rd 2 k 25 k 1.85 k
(e) Av gm(RD rd) (1.88 mS)(1.85 k)
3.48
(f) Av gmRD (1.88 mS)(2 k) 3.76
As demonstrated in part (f), a ratio of 25 k2 k 12.51 between rd and RD
resulted in a difference of 8% in solution.
9.4 JFET SELF-BIAS CONFIGURATION
Bypassed RS
The fixed-bias configuration has the distinct disadvantage of requiring two dc voltage sources. The self-bias configuration of Fig. 9.15 requires only one dc supply to
establish the desired operating point.
412
Chapter 9
FET Small-Signal Analysis
gm
RD
C2
D
C1
Vo
G
Vi
S
Zo
RG
Zi
CS
RS
Figure 9.15 Self-bias JFET configuration.
The capacitor CS across the source resistance assumes its short-circuit equivalence
for dc, allowing RS to define the operating point. Under ac conditions, the capacitor
assumes the short-circuit state and “short circuits” the effects of RS. If left in the ac,
gain will be reduced as will be shown in the paragraphs to follow.
The JFET equivalent circuit is established in Fig. 9.16 and carefully redrawn in
Fig. 9.17.
XC ≈ 0 Ω
XC ≈ 0 Ω
1
2
G
Vi
D
gmVgs
Zi
Vo
rd
RD
RG
S
RS bypassed
by X C
Zo
VDD
S
Figure 9.16 Network of Fig. 9.15 following the substitution of the JFET ac equivalent circuit.
D
G
+
Vi
+
Zi
–
RG
+
gmVgs
Vgs
–
rd
RD
Zo
Vo
–
S
Figure 9.17 Redrawn network of Fig. 9.16.
Since the resulting configuration is the same as appearing in Fig. 9.12, the resulting equations Zi, Zo, and Av will be the same.
Zi:
Zi RG
(9.18)
9.4 JFET Self-Bias Configuration
413
gm
Zo:
Zo rd RD
If rd
(9.19)
10RD,
Zo RD
(9.20)
rd 10RD
Av :
If rd
Av gm(rd RD)
(9.21)
Av gmRD
(9.22)
10RD,
rd 10RD
Phase relationship: The negative sign in the solutions for Av again indicates a
phase shift of 180° between Vi and Vo.
Unbypassed RS
If CS is removed from Fig 9.15, the resistor RS will be part of the ac equivalent circuit as shown in Fig. 9.18. In this case, there is no obvious way to reduce the network to lower its level of complexity. In determining the levels of Zi, Zo, and Av, one
must simply be very careful with notation and defined polarities and direction. Initially, the resistance rd will be left out of the analysis to form a basis for comparison.
D
G
+
+
gmVgs
Zi
Vi
+
ID
Vgs
RG
–
Zo
S
Vo
RD
RS
Io
–
–
Figure 9.18 Self-bias JFET configuration including the effects of RS with rd
.
Zi: Due to the open-circuit condition between the gate and output network, the
input remains the following:
Zi RG
(9.23)
Zo: The output impedance is defined by
Vo
Zo Io
Vi 0
Setting Vi 0 V in Fig. 9.18 will result in the gate terminal being at ground potential (0 V). The voltage across RG is then 0 V, and RG has been effectively “shorted
out” of the picture.
414
Chapter 9
FET Small-Signal Analysis
gm
Applying Kirchhoff’s current law will result in:
ID gmVgS
Io
Vgs (Io
with
so that
ID gm (Io
Io
or
Io ID
ID)RS gmIoRS gmIDRS
gmRS] ID[1
Io[1
and
ID)RS
gmRS]
(the controlled current source gmVgs 0 A
for the applied conditions)
Since
Vo IDRD
then
Vo (Io)RD IoRD
and
Vo
Zo RD
Io
(9.24)
rd ∝
If rd is included in the network, the equivalent will appear as shown in Fig. 9.19.
a
G
+
+
Vi
+
RG
Zo
−
Vgs
−
+
Io
ID
I′
rd
gmVgs
Zi
D
RD
S
Vo
RS
−
−
Io + ID
Io
Figure 9.19 Including the
effects of rd in the self-bias JFET
configuration.
Vo
Zo Io
Since
IDRD
Io
Vi 0 V
we should try to find an expression for Io in terms of ID.
Applying Kirchhoff’s current law:
Io gmVgs
Vrd Vo
but
Io gm
1
IDRD
Vgs ID using Vo IDRD
rd
rd
Vgs (ID
Now,
so that
with the result that
or
Vgs
Vo Vgs
ID
rd
Io gmVgs
and
or
Ird ID
Io gm
1
(ID
rd
I 1
Io) RS
IDRD
Io) RS ID
rd
RS
ID 1
rd
RS
RD
gmRS D
rd
rd
Io RS
1 gmRS rd
Io 1
gmRs
gmRS
RS
rd
RD
rd
9.4 JFET Self-Bias Configuration
415
gm
IDRD
Vo
Zo R
RD
Io
ID 1 gmRs s rd
rd
RS
1 gmRS rd
and
1 g R r
R
R R
1
g
R
r
r RS
m S
and finally,
Zo
d
S
D
d
d
(9.25a)
D
m S
10 RD, 1
For rd
1
gmRS
RS
rd
gmRS
RS
and
rd
RD
and 1
rd
Zo RD
gmRS
RS
rd
RD
rd
(9.25b)
rd
10RD
Av: For the network of Fig. 9.19, an application of Kirchhoff’s voltage law on
the input circuit will result in
Vi Vgs VRS 0
Vgs Vi ID RS
The voltage across rd using Kirchhoff’s voltage law is
Vo VRS
Vo VRS
I rd
and
so that an application of Kirchhoff’s current law will result in
Vo VRS
ID gmVgs rd
Substituting for Vgs from above and substituting for Vo and VRS we have
ID gm[Vi IDRS]
so that
ID 1
gmRS
(IDRD) (IDRS)
rd
RD RS
gmVi
rd
gmVi
ID RD RS
1 gmRS rd
or
The output voltage is then
gmRDVi
Vo IDRD RD RS
1 gmRS rd
and
416
Chapter 9
gmRD
Vo
Av RD RS
Vi
1 gmRS rd
FET Small-Signal Analysis
(9.26)
gm
Again, if rd
10(RD
RS),
gmRD
Vo
Av Vi
1 gmRS
(9.27)
rd 10(RD
RS)
Phase Relationship: The negative sign in Eq. (9.26) again reveals that a 180°
phase shift will exist between Vi and Vo.
The self-bias configuration of Example 6.2 has an operating point defined by VGSQ 2.6 V and IDQ 2.6 mA, with IDSS 8 mA and VP 6 V. The network is redrawn as Fig. 9.20 with an applied signal Vi. The value of yos is given as 20 S.
(a) Determine gm.
(b) Find rd.
(c) Find Zi.
(d) Calculate Zo with and without the effects of rd. Compare the results.
(e) Calculate Av with and without the effects of rd. Compare the results.
EXAMPLE 9.8
20 V
3.3 kΩ
C2
Vo
C1
IDSS = 10 mA
VP = −6 V
Vi
Zo
Zi
1 MΩ
1 kΩ
Figure 9.20 Network for Example 9.8.
Solution
2I SS
2(8 mA)
(a) gm0 D
2.67 mS
VP
6V
VGSQ
(2.6 V)
gm gm0 1 2.67 mS 1 1.51 mS
(6 V)
VP
1
1
(b) rd 50 k
yos
20 S
(c) Zi RG 1 M
(d) With rd:
rd 50 k
10 RD 33 k
Therefore,
Zo RD 3.3 k
If rd Zo RD 3.3 k
(e) With rd:
gmRD
(1.51 mS)(3.3 k)
Av RD RS
3.3 k 1 k
1 gmRS 1 (1.51 mS)(1 k) 50 k
rd
1.92
9.4 JFET Self-Bias Configuration
417
gm
Without rd:
gm RD
(1.51 mS)(3.3 k)
Av 1.98
1 gmRS
1 (1.51 mS)(1 k)
As above, the effect of rd was minimal because the condition rd 10(RD RS)
was satisfied.
Note also that the typical gain of a JFET amplifier is less than that generally encountered for BJTs of similar configurations. Keep in mind, however, that Zi is magnitudes greater than the typical Zi of a BJT, which will have a very positive effect on
the overall gain of a system.
9.5 JFET VOLTAGE-DIVIDER
CONFIGURATION
The popular voltage-divider configuration for BJTs can also be applied to JFETs as
demonstrated in Fig. 9.21.
+VDD
RD
C2
R1
C1
Vo
D
G
Zo
S
Vi
Zi
R2
RS
CS
Figure 9.21 JFET voltage-divider configuration.
Substituting the ac equivalent model for the JFET will result in the configuration
of Fig. 9.22. Replacing the dc supply VDD by a short-circuit equivalent has grounded
one end of R1 and RD. Since each network has a common ground, R1 can be brought
down in parallel with R2 as shown in Fig. 9.23. RD can also be brought down to ground
but in the output circuit across rd. The resulting ac equivalent network now has the
basic format of some of the networks already analyzed.
RD
R1
Vi
Zi
Vo
+
R2
gmVgs
Vgs
RD
Zi
Zo
Zo
R1
R2
gmVgs
Vgs
–
Figure 9.22 Network of Fig. 9.21 under ac conditions.
Chapter 9
Vo
+
–
418
D
G
Vi
Figure 9.23 Redrawn network of Fig. 9.22.
FET Small-Signal Analysis
rd
RD
gm
Zi: R1 and R2 are in parallel with the open-circuit equivalence of the JFET resulting in
Zi R1 R2
(9.28)
Zo: Setting Vi 0 V will set Vgs and gmVgs to zero and
Zo rd RD
For rd
(9.29)
10RD,
Zo RD
(9.30)
rd 10RD
Av :
Vgs Vi
Vo gmVgs(rd RD)
and
gmVgs(rd RD)
Vo
Av Vgs
Vi
so that
Vo
Av gm(rd RD)
Vi
and
If rd
(9.31)
10RD,
Vo
gmRD
Av Vi
(9.32)
rd 10RD
Note that the equations for Zo and Av are the same as obtained for the fixed-bias
and self-bias (with bypassed RS) configurations. The only difference is the equation
for Zi, which is now sensitive to the parallel combination of R1 and R2.
9.6 JFET SOURCE-FOLLOWER
(COMMON-DRAIN) CONFIGURATION
The JFET equivalent of the BJT emitter-follower configuration is the source-follower
configuration of Fig. 9.24. Note that the output is taken off the source terminal and,
when the dc supply is replaced by its short-circuit equivalent, the drain is grounded
(hence, the terminology common-drain).
VDD
D
C1
G
Vi
S
Zi
C2
Vo
RG
RS
Zo
Figure 9.24 JFET source-follower configuration.
9.6 JFET Source-Follower (Common-Drain) Configuration
419
gm
Substituting the JFET equivalent circuit will result in the configuration of Fig.
9.25. The controlled source and internal output impedance of the JFET are tied to
ground at one end and RS on the other, with Vo across RS. Since gmVgs, rd, and RS are
connected to the same terminal and ground, they can all be placed in parallel as shown
in Fig. 9.26. The current source reversed direction but Vgs is still defined between the
gate and source terminals.
G
Vi
D
+
gmVgs
Vgs
Zi
RG
rd
–
S
Vo
Io
S
G
Vi
+
–
Vgs
+
Zi
RS
gmVgs
RG
Zo
Zo
rd
RS
Vo
D
–
Figure 9.25 Network of Fig. 9.24 following the
substitution of the JFET ac equivalent model.
Figure 9.26 Network of Fig. 9.25 redrawn.
Zi: Figure 9.26 clearly reveals that Zi is defined by
Zi RG
(9.33)
Zo: Setting Vi 0 V will result in the gate terminal being connected directly to
ground as shown in Fig. 9.27. The fact that Vgs and Vo are across the same parallel
network results in Vo Vgs.
Io
S
–
+
gmVgs
Vgs
Zo
rd
Vo
RS
–
+
Figure 9.27 Determining Zo for
the network of Fig. 9.24.
Applying Kirchhoff’s current law at node s,
gmVgs Ird
Io
Vo
rd
The result is
1
V r
1
V r
1
Io Vo rd
o
d
o
d
420
Chapter 9
FET Small-Signal Analysis
IRS
Vo
RS
1
g [V ]
R
1
g R
1
gmVgs
RS
m
S
m
S
o
gm
Vo
1
1
Vo
Zo 1
1
1
1
1
1
1
Io
Vo gm
gm
rd
RS
rd
RS
rd
RS
1/gm
which has the same format as the total resistance of three parallel resistors. Therefore,
and
Zo rd RS 1/gm
For rd
(9.34)
10RS,
Zo RS 1/gm
(9.35)
rd 10RS
Av: The output voltage Vo is determined by
Vo gmVgs(rd RS)
and applying Kirchhoff’s voltage law around the perimeter of the network of Fig. 9.26
will result in
Vi Vgs
Vo
Vgs Vi Vo
and
Vo gm(Vi Vo)(rd RS)
so that
or
Vo gmVi(rd RS) gmVo(rd RS)
and
Vo[1
gm(rd RS)] gmVi(rd RS)
gm(rd RS)
Vo
Av Vi
1 gm(rd RS)
so that
In the absence of rd or if rd
(9.36)
10RS,
Vo
gmRS
Av Vi
1 gmRS
(9.37)
rd
10RS
Since the bottom of Eq. (9.36) is larger than the numerator by a factor of one, the
gain can never be equal to or greater than one (as encountered for the emitter-follower BJT network).
Phase Relationship: Since Av of Eq. (9.36) is a positive quantity, Vo and Vi are
in phase for the JFET source-follower configuration.
A dc analysis of the source-follower network of Fig. 9.28 will result in VGSQ 2.86 V
EXAMPLE 9.9
and IDQ 4.56 mA.
(a) Determining gm.
(b) Find rd.
(c) Determine Zi.
+9 V
(d) Calculate Zo with and without rd. Compare results.
IDSS = 16 mA
(e) Determine Av with and without rd. Compare results.
VP = − 4 V
yos = 25 µS
+
0.05 µ F
Vi
Zi
1 MΩ
+
0.05 µ F
2.2 kΩ
–
Zo
Vo
–
Figure 9.28 Network to be analyzed in Example 9.9.
9.6 JFET Source-Follower (Common-Drain) Configuration
421
gm
Solution
2I SS
2(16 mA)
(a) gm0 D
8 mS
VP
4V
VGSQ
(2.86 V)
gm gm0 1 8 mS 1 2.28 mS
VP
(4 V)
1
1
(b) rd 40 k
yos
25 S
(c) Zi RG 1 M
(d) With rd:
Zo rd RS 1/gm 40 k 2.2 k 1/2.28 mS
40 k 2.2 k 438.6 362.52 revealing that Zo is often relatively small and determined primarily by 1/gm. Without rd:
Zo RS 1/gm 2.2 k 438.6 365.69 revealing that rd typically has little impact on Zo.
(e) With rd:
(2.28 mS)(40 k 2.2 k)
gm(rd RS)
Av 1 (2.28 mS)(40 k 2.2 k)
1 gm(rd RS)
(2.28 mS)(2.09 k)
4.77
0.83
1 (2.28 mS)(2.09 k)
1 4.77
which is less than 1 as predicted above.
Without rd:
(2.28 mS)(2.2 k)
gmRS
Av 1 (2.28 mS)(2.2 k)
1 gmRS
5.02
0.83
1 5.02
revealing that rd usually has little impact on the gain of the configuration.
9.7 JFET COMMON-GATE
CONFIGURATION
The last JFET configuration to be analyzed in detail is the common-gate configuration of Fig. 9.29, which parallels the common-base configuration employed with BJT
transistors.
Substituting the JFET equivalent circuit will result in Fig. 9.30. Note the continuing requirement that the controlled source gmVgs be connected from drain to source
with rd in parallel. The isolation between input and output circuits has obviously been
lost since the gate terminal is now connected to the common ground of the network.
In addition, the resistor connected between input terminals is no longer RG but the
resistor RS connected from source to ground. Note also the location of the controlling voltage Vgs and the fact that it appears directly across the resistor RS.
422
Chapter 9
FET Small-Signal Analysis
gm
–
C1
C1
S
C2
D
+
Vi
RD
G
RS
Vi
Zo
C2
b
D
S
Zi
+
gmVgs
RS
Vgs
Z'o
RD
Zo
Vo
Vo
Z'i
VDD
–
+
–
+
+
Zi
a
rd
–
–
+G
–
Figure 9.30 Network of Fig. 9.29 following substitution of
JFET ac equivalent model.
Figure 9.29 JFET common-gate configuration.
Zi: The resistor RS is directly across the terminals defining Zi. Let us therefore
find the impedance Zi of Fig. 9.29, which will simply be in parallel with RS when Zi
is defined.
The network of interest is redrawn as Fig. 9.31. The voltage V Vgs. Applying Kirchhoff’s voltage law around the output perimeter of the network will result in
V Vrd VRD 0
Vrd V VRD V IRD
and
I'
I'
–
+
a
+
Ird
Z'i
rd V r d
gmVgs
Vi
Vgs
–
+
–
+
RD VRD
I'
–
Figure 9.31 Determining Zi for
the network of Fig. 9.29.
Applying Kirchhoff’s current law at node a results in
I
and
gmVgs Ird
(V IRD)
I Ird gmVgs gmVgs
rd
or
V
IRD
I gm[V]
rd
rd
so that
I 1
RD
1
V rd
rd
gm
1 r
V Z 1
I
g r
RD
d
and
i
(9.38)
m
d
or
V
rd RD
Zi I
1 gmrd
9.7 JFET Common-Gate Configuration
423
gm
Zi RS Zi
and
rd RD
Zi RS 1 gmrd
results in
(9.39)
If rd 10RD, Eq. (9.38) permits the following approximation since RD /rd 1 and
1/rd gm:
1 r 1
Z 1 g
g
r
RD
d
i
m
m
d
Zi RS 1/gm
and
(9.40)
rd 10RD
Zo: Substituting Vi 0 V in Fig. 9.30 will “short-out” the effects of RS and set
Vgs to 0 V. The result is gmVgs 0, and rd will be in parallel with RD. Therefore,
Zo RD rd
For rd
(9.41)
10RD,
Zo RD
(9.42)
rd 10RD
Av: Figure 9.30 reveals that
Vi Vgs
Vo IDRD
and
The voltage across rd is
Vrd Vo Vi
Vo Vi
Ird rd
and
Applying Kirchhoff’s current law at node b in Fig. 9.30 results in
ID
Ird
gmVgs 0
ID Ird gmVgs
and
Vo Vi
gm[Vi]
rd
Vi Vo
ID rd
so that
gmVi
Vi Vo
Vo IDRD rd
V RD
V RD
i o rd
rd
and
424
Chapter 9
Vo 1
RD
RD
Vi rd
rd
FET Small-Signal Analysis
gmVi RD
gm
gmRD
gm
g R
r V
A R
V
1 r
RD
m D
d
o
with
v
(9.43)
D
i
d
For rd 10RD, the factor RD /rd of Eq. (9.43) can be dropped as a good approximation and
Av gmRD
(9.44)
rd 10RD
Phase Relationship: The fact that Av is a positive number will result in an inphase relationship between Vo and Vi for the common-gate configuration.
Although the network of Fig. 9.32 may not initially appear to be of the common-gate
variety, a close examination will reveal that it has all the characteristics of Fig. 9.29.
If VGSQ 2.2 V and IDQ 2.03 mA:
(a) Determine gm.
(b) Find rd.
(c) Calculate Zi with and without rd. Compare results.
(d) Find Zo with and without rd. Compare results.
(e) Determine Vo with and without rd. Compare results.
EXAMPLE 9.10
+12 V
3.6 kΩ
10 µ F
Vo
IDSS = 10 mA
VP = − 4 V
yos = 50 µS
10 µ F
+
Vi = 40 mV
1.1 kΩ
–
Figure 9.32 Network for Example 9.10.
Solution
2I SS
2(10 mA)
(a) gm0 D
5 mS
VP
4V
VGSQ
(2.2 V)
5 mS 1 2.25 mS
gm gm0 1 VP
(4 V)
1
1
(b) rd 20 k
yos
50 S
(c) With rd:
20 k 3.6 k
rd RD
Zi RS 1.1 k 1 (2.25 ms)(20 k)
1 gmrd
1.1 k 0.51 k 0.35 k
9.7 JFET Common-Gate Configuration
425
gm
Without rd:
Zi RS 1/gm 1.1 k 1/2.25 ms 1.1 k 0.44 k
0.31 k
Even though the condition,
rd
10RD 20 k
10(3.6 k) 20 k
36 k
is not satisfied, both equations result in essentially the same level of impedance.
In this case, 1/gm was the predominant factor.
(d) With rd:
Zo RD rd 3.6 k 20 k 3.05 k
Without rd:
Zo RD 3.6 k
Again the condition rd 10RD is not satisfied, but both results are reasonably
close. RD is certainly the predominant factor in this example.
(e) With rd:
RD
3.6 k
gmRD (2.25 mS)(3.6 k) rd
20 k Av RD
3.6 k
1 1 rd
20 k 8.1 0.18
7.02
1 0.18
Vo
Av ➤ Vo AvVi (7.02)(40 mV) 280.8 mV
Vi
and
Without rd:
Av gmRD (2.25 mS)(3.6 k) 8.1
with
Vo AvVi (8.1)(40 mV) 324 mV
In this case, the difference is a little more noticeable but not dramatically so.
Example 9.10 demonstrates that even though the condition rd 10RD was not satisfied, the results for the parameters given were not significantly different using the
exact and approximate equations. In fact, in most cases, the approximate equations
can be used to find a reasonable idea of particular levels with a reduced amount of
effort.
9.8 DEPLETION-TYPE MOSFETs
The fact that Shockley’s equation is also applicable to depletion-type MOSFETs results in the same equation for gm. In fact, the ac equivalent model for D-MOSFETs
is exactly the same as that employed for JFETs as shown in Fig. 9.33.
The only difference offered by D-MOSFETs is that VGSQ can be positive for
n-channel devices and negative for p-channel units. The result is that gm can be greater
than gm0 as demonstrated by the example to follow. The range of rd is very similar to
that encountered for JFETs.
426
Chapter 9
FET Small-Signal Analysis
gm
G
+
D
G
D
Vgs
G
S
gmVgs
rd
–
S
S
Figure 9.33 D-MOSFET ac equivalent model.
The network of Fig. 9.34 was analyzed as Example 6.8, resulting in VGSQ 0.35 V
and IDQ 7.6 mA.
(a) Determine gm and compare to gm0.
(b) Find rd.
(c) Sketch the ac equivalent network for Fig. 9.34.
(d) Find Zi.
18 V
(e) Calculate Zo.
(f) Find Av.
EXAMPLE 9.11
1.8 kΩ
C2
110 MΩ
Vo
IDSS = 6 mA
VP = − 3 V
yos = 10 µS
C1
Vi
Zi
150 Ω
10 MΩ
Zo
Figure 9.34 Network for Example 9.11.
Solution
2I SS
2(6 mA)
(a) gm0 D
4 mS
VP
3V
VGSQ
( 0.35 V)
4 mS 1 4 mS(1 0.117) 4.47 mS
gm gm0 1 VP
( 3 V)
1
1
(b) rd 100 k
yos
10 S
(c) See Fig. 9.35. Note the similarities with the network of Fig. 9.23. Equations (9.28)
through (9.32) are therefore applicable.
G
+
Vi
D
+
Zi
10 MΩ
110 MΩ
–
+
4.47 × 10 −3 Vgs
Vgs
100 kΩ
Zo
Vo
–
–
S
1.8 kΩ
S
Figure 9.35 AC equivalent circuit for Fig. 9.34.
9.8 Depletion-Type MOSFETs
427
gm
(d) Eq. (9.28): Zi R1 R2 10 M 110 M 9.17 M
(e) Eq. (9.29): Zo rd RD 100 k 1.8 k 1.77 k RD 1.8 k
(f) rd 10RD → 100 k 18 k
Eq. (9.32): Av gmRD (4.47 mS)(1.8 k) 8.05
9.9 ENHANCEMENT-TYPE MOSFETs
The enhancement-type MOSFET can be either an n-channel (nMOS) or p-channel
(pMOS) device, as shown in Fig. 9.36. The ac small-signal equivalent circuit of either device is shown in Fig. 9.36, revealing an open-circuit between gate and drainsource channel and a current source from drain to source having a magnitude dependent on the gate-to-source voltage. There is an output impedance from drain to source
rd, which is usually provided on specification sheets as an admittance yos. The device
transconductance, gm, is provided on specification sheets as the forward transfer admittance, yfs.
D
pMOS
G
D
G
S
+
Vgs
D
gmVgs
rd
–
S
nMOS
G
S
gm =yfs , rd = 1
yos
Figure 9.36 Enhancement MOSFET ac small-signal model.
In our analysis of JFETs, an equation for gm was derived from Shockley’s equation. For E-MOSFETs, the relationship between output current and controlling voltage is defined by
ID k(VGS VGS(Th))2
Since gm is still defined by
ID
gm VGS
we can take the derivative of the transfer equation to determine gm as an operating
point. That is,
dID
d
d
k(VGS VGS(Th))2 k (VGS VGS(Th))2
gm dVGS
dVGS
dVGS
d
2k(VGS VGS(Th)) (VGS VGS(Th)) 2k(VGS VGS(Th))(1 0)
dVGS
and
428
Chapter 9
gm 2k(VGSQ VGS(Th))
FET Small-Signal Analysis
(9.45)
gm
Recall that the constant k can be determined from a given typical operating point on
a specification sheet. In every other respect, the ac analysis is the same as that employed for JFETs or D-MOSFETs. Be aware, however, that the characteristics of an
E-MOSFET are such that the biasing arrangements are somewhat limited.
9.10 E-MOSFET DRAIN-FEEDBACK
CONFIGURATION
The E-MOSFET drain-feedback configuration appears in Fig. 9.37. Recall from dc
calculations that RG could be replaced by a short-circuit equivalent since IG 0 A
and therefore VRG 0 V. However, for ac situations it provides an important high impedance between Vo and Vi. Otherwise, the input and output terminals would be connected directly and Vo Vi.
VDD
RD
RF
C2
RF
Vo
Zo
C1
Vi
Ii
Ii
D
+
G
Zi
Vi
G
D
Vgs
–
S
gmVgs
rd
–
Zi
Figure 9.37 E-MOSFET
drain-feedback configuration.
Vo
+
RD
Zo
S
Figure 9.38 AC equivalent of the
network of Fig. 9.37.
Substituting the ac equivalent model for the device will result in the network of
Fig. 9.38. Note that RF is not within the shaded area defining the equivalent model
of the device but does provide a direct connection between input and output circuits.
Zi: Applying Kirchhoff’s current law to the output circuit (at node D in Fig.
9.38) results in
Vo
Ii gmVgs rd RD
Vgs Vi
and
Vo
rd RD
so that
Ii gmVi
or
Vo
Ii gmVi rd RD
Therefore,
with
Vo (rd RD)(Ii gmVi)
Vi (rd RD)(Ii gmVi)
Vi Vo
Ii RF
RF
and
IiRF Vi (rd RD)Ii
so that
Vi[1
and finally,
(rd RD)gmVi
gm(rd RD)] Ii[RF
rd RD]
RF rd RD
V
Zi i 1 gm(rd RD)
Ii
(9.46)
9.10 E-MOSFET Drain-Feedback Configuration
429
gm
Typically, RF
rd RD, so that
RF
Zi 1 gm(rd RD)
For rd
10RD,
RF
Zi 1 gmRD
(9.47)
RF
rd RD, rd
10RD
Zo: Substituting Vi 0 V will result in Vgs 0 V and gmVgs 0, with a shortcircuit path from gate to ground as shown in Fig. 9.39. RF, rd, and RD are then in
parallel and
Zo RF rd RD
(9.48)
RF
gmVgs = 0 mA
Vi = Vgs = 0 V
rd
RD
Zo
Figure 9.39 Determining Zo for the network of Fig. 9.37.
Normally, RF is so much larger than rd RD that
Zo rd RD
and with rd
10RD,
Zo RD
(9.49)
RF
rd RD, rd
10RD
Av: Applying Kirchhoff’s current law at node D of Fig. 9.38 will result in
Ii gmVgs
but
Vi Vo
Vgs Vi and Ii RF
so that
Vi Vo
gmVi
RF
Vo
rd RD
and
V
Vo
i gmVi
RF
RF
Vo
rd RD
so that
and
430
Vo
rd RD
Chapter 9
1
Vo rd RD
1
1
Vi gm
RF
RF
1
gm
RF
Vo Av 1 1
Vi
rd RD RF
FET Small-Signal Analysis
1
rd RD
but
and
1
RF
gm
Av gm(RF rd RD)
so that
Since RF is usually
gm
1
1
RF
RF rd RD
rd RD and if rd
(9.50)
10RD,
Av gmRD
(9.51)
RF
rd RD, rd 10RD
Phase Relationship: The negative sign for Av reveals that Vo and Vi are out of
phase by 180°.
The E-MOSFET of Fig. 9.40 was analyzed in Example 6.11, with the result that
k 0.24 103 A/V2, VGSQ 6.4 V, and IDQ 2.75 mA.
(a) Determine gm.
(b) Find rd.
(c) Calculate Zi with and without rd. Compare results.
(d) Find Zo with and without rd. Compare results.
(e) Find Av with and without rd. Compare results.
EXAMPLE 9.12
12 V
2 kΩ
Vo
1 µF
10 MΩ
Vi
1 µF
Zo
ID ( on) = 6 mA
VGS (on) = 8 V
VGS (Th) = 3 V
yos = 20 µS
Zi
Figure 9.40 Drain-feedback amplifier from Example 6.11.
Solution
(a) gm 2k(VGSQ VGS(Th)) 2(0.24 103 A/V2)(6.4 V 3 V)
1.63 mS
1
1
(b) rd 50 k
yos
20 S
(c) With rd:
10 M 50 k 2 k
RF rd RD
Zi 1
(1.63 mS)(50 k 2 k)
1 gm(rd RD)
10 M 1.92 k
2.42 M
1 3.13
9.10 E-MOSFET Drain-Feedback Configuration
431
gm
Without rd:
RF
10 M
Zi 2.53 M
1 (1.63 mS)(2 k)
1 gmRD
revealing that since the condition rd 10RD 50 k
results for Zo with or without rd will be quite close.
(d) With rd:
40 k is satisfied, the
Zo RF rd RD 10 M 50 k 2 k 49.75 k 2 k
1.92 k
Without rd:
Zo RD 2 k
again providing very close results.
(e) With rd:
Av gm(RF rd RD)
(1.63 mS)(10 M 50 k 2 k)
(1.63 mS)(1.92 k)
3.21
Without rd:
Av gmRD (1.63 mS)(2 k)
3.26
which is very close to the above result.
9.11 E-MOSFET VOLTAGE-DIVIDER
CONFIGURATION
The last E-MOSFET configuration to be examined in detail is the voltage-divider network of Fig. 9.41. The format is exactly the same as appearing in a number of earlier discussions.
Substituting the ac equivalent network for the E-MOSFET will result in the configuration of Fig. 9.42, which is exactly the same as Fig. 9.23. The result is that Eqs.
(9.28) through (9.32) are applicable as listed below for the E-MOSFET.
VDD
RD
Vo
R1
D
C1
Zo
G
Vi
Vi
S
Zi
R2
RS
CS
D
G
Vo
+
Zi
R1
R2
Zo
Vgs
gmVgs
rd
RD
–
S
Figure 9.41 E-MOSFET
voltage-divider configuration.
432
Figure 9.42 AC equivalent network for the configuration of Fig. 9.41.
Chapter 9
FET Small-Signal Analysis
gm
Zi:
Zi R1 R2
(9.52)
Zo rd RD
(9.53)
Zo:
For rd
10RD,
Zo RD
rd 10RD
(9.54)
Av :
and if rd
Vo
Av gm(rD RD)
Vi
(9.55)
Vo
gmRD
Av Vi
(9.56)
10RD,
9.12 DESIGNING FET
AMPLIFIER NETWORKS
Design problems at this stage are limited to obtaining a desired dc bias condition or
ac voltage gain. In most cases, the various equations developed are used “in reverse”
to define the parameters necessary to obtain the desired gain, input impedance, or
output impedance. To avoid unnecessary complexity during the initial stages of the
design, the approximate equations are often employed because some variation will
occur when calculated resistors are replaced by standard values. Once the initial design is completed, the results can be tested and refinements made using the complete
equations.
Throughout the design procedure be aware that although superposition permits a
separate analysis and design of the network from a dc and an ac viewpoint, a parameter chosen in the dc environment will often play an important role in the ac response. In particular, recall that the resistance RG could be replaced by a short-circuit
equivalent in the feedback configuration because IG 0 A for dc conditions, but
for the ac analysis, it presents an important high impedance path between Vo and
Vi. In addition, recall that gm is larger for operating points closer to the ID axis
(VGS 0 V), requiring that RS be relatively small. In the unbypassed RS network, a
small RS will also contribute to a higher gain, but for the source-follower, the gain is
reduced from its maximum value of 1. In total, simply keep in mind that network parameters can affect the dc and ac levels in different ways. Often a balance must be
made between a particular operating point and its impact on the ac response.
In most situations, the available dc supply voltage is known, the FET to be employed has been determined, and the capacitors to be employed at the chosen frequency are defined. It is then necessary to determine the resistive elements necessary
to establish the desired gain or impedance level. The next three examples will determine the required parameters for a specific gain.
9.12 Designing FET Amplifier Networks
433
gm
EXAMPLE 9.13
Design the fixed-bias network of Fig. 9.43 to have an ac gain of 10. That is, determine the value of RD.
VDD (+30 V)
RD
Vo
C1
Vi
0.1 µF
RG
10 MΩ
IDSS = 10 mA
VP = – 4 V
yos = 20 µS
Figure 9.43 Circuit for desired
voltage gain in Example 9.13.
Solution
Since VGSQ 0 V, the level of gm is gm0. The gain is therefore determined by
Av gm(RD rd) gm0(RD rd)
2I SS
2(10 mA)
gm0 D
5 mS
VP
4V
with
The result is
10 5 mS(RD rd)
and
10
RD rd 2 k
5 mS
From the device specifications,
1
1
rd 50 k
yos
20 106 S
Substituting, we find
RD rd RD 50 k 2 k
and
or
RD(50 k)
2 k
RD 50 k
50RD 2(RD
50 k) 2RD
with
48RD 100 k
and
100 k
RD 2.08 k
48
100 k
The closest standard value is 2 k (Appendix C), which would be employed for this
design.
The resulting level of VDSQ would then be determined as follows:
VDSQ VDD IDQRD 30 V (10 mA)(2 k) 10 V
The levels of Zi and Zo are set by the levels of RG and RD, respectively. That is,
Zi RG 10 M
Zo RD rd 2 k 50 k 1.92 k RD 2 k.
434
Chapter 9
FET Small-Signal Analysis
gm
Choose the values of RD and RS for the network of Fig. 9.44 that will result in a gain
of 8 using a relatively high level of gm for this device defined at VGSQ 14VP.
EXAMPLE 9.14
VDD
+20 V
RD
C2
Vo
C1
Vi
0.1 µF
0V
RL
10 MΩ
0.1 µF
RG
10 MΩ
CS
40 µF
RS
IDSS = 10 mA
VP = – 4 V
yos = 20 µS
gm 0 = 5 mS
Figure 9.44 Network for desired voltage gain in Example 9.14.
Solution
The operating point is defined by
1
1
VGSQ VP (4 V) 1 V
4
4
and
VGSQ
ID IDSS 1 VP
(1 V)
10 mA 1 (4 V)
2
2
5.625 mA
Determining gm,
VGSQ
gm gm0 1 VP
(1 V)
5 mS 1 3.75 mS
(4 V)
The magnitude of the ac voltage gain is determined by
Av gm(RD rd)
Substituting known values will result in
8 (3.75 mS)(RD rd)
so that
8
RD rd 2.13 k
3.75 mS
The level of rd is defined by
1
1
rd 50 k
yos
20 S
and
RD 50 k 2.13 k
with the result that
RD 2.2 k
which is a standard value.
9.12 Designing FET Amplifier Networks
435
gm
The level of RS is determined by the dc operating conditions as follows:
VGSQ IDRS
1 V (5.625 mA)RS
1V
RS 177.8 5.625 mA
and
The closest standard value is 180 . In this example, RS does not appear in the ac
design because of the shorting effect of CS.
In the next example, RS is unbypassed and the design becomes a bit more complicated.
EXAMPLE 9.15
Determine RD and RS for the network of Fig. 9.44 to establish a gain of 8 if the bypass capacitor CS is removed.
Solution
VGSQ and IDQ are still 1 V and 5.625 mA, and since the equation VGS IDRS has
not changed, RS continues to equal the standard value of 180 obtained in Example 9.14.
The gain of an unbypassed self-bias configuration is
gmRD
Av 1 gmRS
For the moment it is assumed that rd 10(RD RS). Using the full equation for
Av at this stage of the design would simply complicate the process unnecessarily.
Substituting (for the specified magnitude of 8 for the gain),
(3.75 mS)RD
(3.75 mS)RD
8 1 (3.75 mS)(180 )
1 0.675
0.675) (3.75 mS)RD
and
8(1
so that
13.4
RD 3.573 k
3.75 mS
with the closest standard value at 3.6 k.
We can now test the condition:
rd
and
10(RD
RS)
50 k
10(3.6 k
50 k
37.8 k
0.18 k) 10(3.78 k)
which is satisfied—the solution stands!
9.13 SUMMARY TABLE
In an effort to provide a quick comparison between configurations and offer a listing
that can be helpful for a variety of reasons, Table 9.1 was developed. The exact and
approximate equation for each important parameter are provided with a typical range
of values for each. Although all the possible configurations are not present, the majority of the most frequently encountered are included. In fact, any configuration not
436
Chapter 9
FET Small-Signal Analysis
gm
TABLE 9.1 Zi, Z0, and Av for various FET configurations
Configuration
Zi
Vo
Av Vi
Zo
Fixed-bias
[JFET or D-MOSFET]
+VDD
Medium (2 k)
D-MOSFET]
RD
C2
Vo
C1
Vi
High (10 M)
Zo
RD rd
RD
gm(rd RD)
gm RD
RG
RG
Zi
Medium (10)
(rd
(rd
10 RD)
10 RD)
VGG
Self-bias
bypassed Rs
[JFET or D-MOSFET]
+VDD
Medium (2 k)
Medium (10)
S
-MOSFET]
RD
High (10 M)
C2
Vo
C1
Vi
RD rd
RD
(rd
RG
RS
gm(rd RD)
gm RD
RG
Zo
Zi
(rd
10 RD)
10 RD)
CS
Self-bias
unbypassed RS
[JFET or D-MOSFET]
Low (2)
+VDD
d RS
-MOSFET]
RD
High (10 M)
C2
Vo
C1
Vi
RG
RD
gmRD
RD RS
1 gmRs rd
RG
Zo
Zi
R
1 gmRS S RD
rd
R
RD
1 gmRS S rd
rd
rd
10 RD or rd gmRD
1 gmRS
[rd
RS
10(Rd
RS)]
Voltage-divider bias
[JFET or D-MOSFET]
+VDD
vider bias
-MOSFET]
RD
R1
C2
Vo
C1
Vi
Zo
Zi
R2
High (10 M)
Medium (2 k)
Medium (10)
RD r
gm(rd RD)
RD
gm RD
R1 R2
(rd
RS
10 RD)
(rd
10 RD)
CS
9.13 Summary Table
437
gm
TABLE 9.1 (Continued)
Configuration
Zi
V
Av Vo
Zo
i
Source-follower
[JFET or D-MOSFET]
Low ( 1)
Low (100 k)
+VDD
-MOSFET]
C1
High (10 M)
Vi
C2
Zi
Vo
RG
RS
RG
rd RS 1/gm
RS 1/gm
(rd
10 RS)
Zo
gm(rd RS)
1 gm(rd RS)
gmRS
1 gmRS
(rd
10 R S)
Common-gate
[JFET or D-MOSFET]
Medium (2 k)
+VDD
ate
-MOSFET]
Q1
C1
RD
C2
rd RD
RS 1 gmrd
1
RS gm
Vo
Vi
RS
Zi
RG
Medium ( 10)
RD
gmRD rd
R
D
1 rd
Low (1 k)
CS
Zo
(rd
RD rd
RD
(rd
10 RD)
10 RD)
gmRD
(rd
10 R S)
Drain-feedback bias
E-MOSFET
Medium (1 M)
+VDD
back bias
T
RD
RF
Medium (2 k)
C2
RF rd RD
1 gm(rd RD)
RF
1 gmRD
RF rd RD
RD
Medium (10)
gm(RF rd RD)
gmRD
Vo
C1
Vi
Zo
(rd
(RF, rd
10 RD)
(RF, rd
10 RD)
10 RD)
Zi
Voltage-divider bias
E-MOSFET
ider bias
T
+VDD
RD
R1
C1
Vi
C2
Medium (1 M)
D
Vo
G
Zo
S
Zi
438
R2
Medium (2 k)
Medium (10)
RD rd
gm(rd RD)
RD
gmRD
R1 R2
RS
Chapter 9
FET Small-Signal Analysis
(Rd
10 RD)
(rd
10 RD)
gm
listed will probably be some variation of those appearing in the table, so at the very
least, the listing will provide some insight as to what expected levels should be and
which path will probably generate the desired equations. The format chosen was designed to permit a duplication of the entire table on the front and back of one 812 by
11 inch page.
9.14 TROUBLESHOOTING
As mentioned before, troubleshooting a circuit is a combination of knowing the theory and having experience using meters and an oscilloscope to check the operation
of the circuit. A good troubleshooter has a “nose” for finding the trouble in a
circuit—this ability to “see” what is happening being greatly developed through building, testing, and repairing many different circuits. For an FET small-signal amplifier,
one could go about troubleshooting a circuit by performing a number of basic steps:
1. Look at the circuit board to see if any obvious problems can be seen: an area
charred by excess heating of a component; a component that feels or seems too
hot to touch; what appears to be a poor solder joint; any connection that appears
to have come loose.
2. Use a dc meter: make some measurements as marked in a repair manual containing the circuit schematic diagram and a listing of test dc voltages.
3. Apply a test ac signal: measure the ac voltages starting at the input and working
along toward the output.
4. If the problem is identified at a particular stage, the ac signal at various points
should be checked using an oscilloscope to see the waveform, its polarity, amplitude, and frequency, as well as any unusual waveform “glitches” that may be present. In particular, observe that the signal is present for the full signal cycle.
Possible Symptoms and Actions
If there is no output ac voltage:
1.
2.
3.
4.
Check
Check
Check
Check
if the supply voltage is present.
if the output voltage at VD is between 0 V and VDD.
if there is any input ac signal at the gate terminal.
the ac voltage at each side of the coupling capacitor terminals.
When building and testing a FET amplifier circuit in the laboratory:
1. Check the color code of resistor values to be sure that they are correct. Even better, measure the resistor value as components used repeatedly may get overheated
when used incorrectly, causing the nominal value to change.
2. Check that all dc voltages are present at the component terminals. Be sure that all
ground connections are made common.
3. Measure the ac input signal to be sure the expected value is provided to the circuit.
9.15 PSPICE WINDOWS
JFET Fixed-Bias Configuration
The first JFET configuration to be analyzed using PSpice Windows is the fixed-bias
configuration of Fig. 9.45, which has a JFET with VP 4 V and IDSS 10 mA. The 10-M resistor was added to act as a path to ground for the capacitor
9.15 PSpice Windows
439
gm
but is essentially an open-circuit as a load. The J2N3819 n-channel JFET from the
EVAL.slb library will be used, and the ac voltage will be determined at four different points for comparison and review.
Figure 9.45 Fixed-bias JFET
configuration with an ac source.
The constant Beta is determined by
10 mA
I SS
Beta D
0.625 mA/V2
42
Vp 2
and inserted as a Model Parameter using the sequence Edit-Model-Edit Instance
Model (Text). Vto must also be changed to 4 V. The remaining elements of the network are set as described for the transistor in Chapter 8.
An analysis of the network will result in the printout of Fig. 9.46. The Schematics Netlist reveals the nodes assigned to each parameter and defines the nodes for
which the ac voltage is to be printed. In this case, note that Vi is set at 10 mV at a
frequency of 10kHz from node 2 to 0. In the list of Junction FET MODEL
PARAMETERS, VTO is 4 V and BETA is 625E-6 as entered. The SMALLSIGNAL BIAS SOLUTION reveals that the voltage at both ends of RG is 1.5 V,
resulting in VGS 1.5 V. The voltage from drain to source (ground) is 12 V, leaving a drop of 8 V across RD. The AC ANALYSIS at the end of the listing reveals that
the voltage at the source (node 2) is 10 mV as set, but the voltage at the other end of
the capacitor is 3 V less due to the impedance of the capacitor at 10 kHz—certainly
a drop to be ignored. The choice of 0.02 F for this frequency was obviously a good
one. The voltages before and after the capacitor on the output side are exactly the
same (to three places), revealing that the larger the capacitor, the closer the characteristics to a short circuit. The output of 6.275E-2 62.75 mV reflects a gain of 6.275.
The OPERATING POINT INFORMATION reveals that ID is 4 mA and gm is
3.2 mS. Calculating the value of gm from:
VGSQ
2I SS
gm D
1 VP
VP
2 (10 mA)
(1.5 V)
1 4V
(4 V)
3.125 mS
confirming our analysis.
440
Chapter 9
FET Small-Signal Analysis
gm
Figure 9.46 Output file for the
network of Figure 9.45.
9.15 PSpice Windows
441
gm
JFET Self-Bias Configuration
The self-bias configuration of Fig. 9.47 will be analyzed using the J2N3819 JFET
from the library and then using an approximate equivalent circuit. It will be interesting to see if there are any major differences in solution.
Figure 9.47 Self-bias
configuration with an ac source.
Again, VP 4 V and IDSS 10 mA, resulting in a Vto of 4 and a Beta of
6.25E-4. The Analysis is run and the results of Fig. 9.48 obtained. The nodes are
identified in the Schematics Netlist and the parameters in the Junction FET MODEL
PARAMETERS. The SMALL-SIGNAL BIAS SOLUTION reveals that VGS 1.7114 V and VD 14.228 V—results that are very close to a hand-written solution of 1.68 V and 14.49 V. The OPERATING POINT INFORMATION reveals
that ID is 3.36 mA compared to a hand-calculated level of 3.3 mA and that gm is 2.94
mS compared to a hand-calculated level of 2.90 mS. The AC ANALYSIS provides
an output level of 13.3 mV at an angle of 179.9°, which compares well with a handcalculated level of 13.63 mV at an angle of 180°. The results for JFETs are a lot
closer than those obtained for transistors when we used the provided elements because of the special feature of having essentially infinite input impedance so that the
gate current is zero ampere. Recall that for the transistor, VBE is a function of the operating conditions.
We will now investigate the self-bias configuration using the approximate model
as done for the transistor and see if there is an improvement in the results (compared
to the hand-calculated levels). In this case, we need the voltage controlled current
source (VCCS) found in the ANALOG.slb library as G. When selected, the Description reads Voltage-controlled current source. When placed on the schematic,
it will appear as shown in Fig. 9.49. The sensing voltage is between the plus and minus sign, while the controlled current is between the other two external terminals.
Figure 9.49 Network in
Figure 9.47 following substitution
of a VCCS for the JFET in the ac
domain.
442
Chapter 9
FET Small-Signal Analysis
gm
Figure 9.48 Output file for the
network of Figure 9.47.
9.15 PSpice Windows
443
gm
Figure 9.50 JFET voltage-divider
configuration with an ac source.
Double-clicking on the schematic symbol will result in a PartName: G dialog box,
in which the GAIN(gm) can be set to the hand-calculated level of 2.90 mS.
The result of an analysis is a gain of 13.62—almost an exact match with the handwritten gain. This approach is certainly valid for an ac analysis, but if we examine
the SMALL-SIGNAL BIAS SOLUTION, we find that the results are meaningless.
Therefore, the equivalent appearing in Fig. 9.49 is only valid for the ac gain since the
only parameter defined is the ac transconductance factor.
JFET Voltage-Divider Configuration
The last network to be analyzed in this PSpice Windows presentation is the voltagedivider configuration of Fig. 9.50. Note that the parameters chosen are different from
those employed in earlier examples, with Vi at 24 mV and a frequency of 5 kHz. In
addition, the dc levels are displayed and a plot of the output and input voltages will
be obtained on the same screen.
After setting up the network, the source Vi must be set to the indicated parameters by double-clicking on the source and then sequentially double-clicking on each
parameter and typing in the correct values. Each must be saved and then the display
changed to print the magnitude of the ac voltage and the applied frequency. In this
example, the JFET parameters were printed on the screen using the ABC icon. BETA
is of course calculated from IDSS/VP2. Under Analysis-Probe Setup, the option Do
not auto-run Probe was chosen, and under Setup, AC Sweep was chosen and the
frequency of 5 kHz entered. Finally, since we want the dc levels to be displayed, the
Display Results on Schematic option is chosen under Analysis, and Enable
Voltage Display is enabled. The resulting dc levels of Fig. 9.50 reveal that VGS is
1.823 V 3.635 V 1.812 V, comparing very well with the 1.8 V calculated in
Example 6.5. VD is 10.18 V compared to the calculated level of 10.24 V, and VDS is
10.18 V 3.635 V 6.545 V compared to 6.64 V.
For the ac solution, we can choose Examine Output under Analysis and find under OPERATING POINT INFORMATION that gm is 2.22 mS, comparing very
well with the hand-calculated value of 2.2 mS, and under AC ANALYSIS that the
output ac voltage is 125.8 mV, resulting in a gain of 125.8 mV/24 mV 5.24. The
hand-calculated level is gmRD (2.2 mS)(2.4 k) 5.28. The ac waveform for the
output can be obtained by first applying the sequence Analysis-Probe SetupAutomatically run Probe after simulation. Then, return to Setup under Analysis,
and enable Transient, disable AC Sweep, and double-click Transient to obtain the
Transient dialog box. For the frequency of 5 kHz, the period is 200 s. A Print Step
to 2 s would then give us 100 plot points for each cycle. The Final Time will be
5 200 s 1 ms to show five cycles. The No-Print Delay will be 0s and the Step
Ceiling 2s. Then, click the Trace icon, choose V(J1:d), and the output waveform
of Fig. 9.51 will appear. Choose Plot-Add Plot-Trace-Add-V(Vi:+), and both wave444
Chapter 9
FET Small-Signal Analysis
gm
forms will appear as shown. Shift SEL
to the bottom waveform by simply bringing the pointer to the left of the lower waveform and left-clicking the mouse once.
Click the Toggle cursor icon, and a horizontal line will appear at the dc level of the
output voltage at 10.184 V. A left click of the mouse and an intersecting set of lines
will appear. Choose the Cursor Peak icon, and the intersection will automatically go
to the peak value of the waveform (A1 in the dialog box). The difference appearing
in the dialog box is 125.496 mV, comparing well with the printed value in the output
file. The difference is simply due to the number of points chosen for the plot; an increased number of plot points would have brought the two levels closer together.
Figure 9.51 The ac drain and gate voltage for the voltage-divider JFET
configuration of Figure 9.50.
PROBLEMS
§ 9.2 FET Small-Signal Model
1. Calculate gm0 for a JFET having device parameters IDSS 15 mA and VP 5 V.
2. Determine the pinch-off voltage of a JFET with gm0 10 mS and IDSS 12 mA.
3. For a JFET having device parameters gm0 5 mS and VP 3.5 V, what is the device current at VGS 0 V?
4. Calculate the value of gm for a JFET (IDSS 12 mA, VP 3 V) at a bias point of VGS 1 V.
5. For a JFET having gm 6 mS at VGSQ 1 V, what is the value of IDSS if VP 2.5 V?
6. A JFET (IDSS 10 mA, VP 5 V) is biased at ID IDSS/4. What is the value of gm at that
bias point?
7. Determine the value of gm for a JFET (IDSS 8 mA, VP 5 V) when biased at VGSQ VP/4.
8. A specification sheet provides the following data (at a listed drain–source current)
yfs 4.5 mS,
yos 25 S
At the listed drain–source current, determine:
(a) gm.
(b) rd.
9. For a JFET having specified values of yfs 4.5 mS and yos 25 S, determine the device output impedance, Zo(FET), and device ideal voltage gain, Av(FET).
Problems
445
gm
10. If a JFET having a specified value of rd 100 k has an ideal voltage gain of Av(FET) 200, what is the value of gm?
11. Using the transfer characteristic of Fig. 9.52:
(a) What is the value of gm0?
(b) Determine gm at VGS 1.5 V graphically.
(c) What is the value of gm at VGSQ 1.5 V using Eq. (9.6)? Compare with the solution to
part (b).
(d) Graphically determine gm at VGS 2.5 V.
(e) What is the value of gm at VGSQ 2.5 V using Eq. (9.6)? Compare with the solution to
part (d).
ID (mA)
10
9
8
7
6
5
4
3
2
1
−5
−4
−3
−2
−1
0
VGS (V)
Figure 9.52 JFET transfer
characteristic for Problem 11
12. Using the drain characteristic of Fig. 9.53:
(a) What is the value of rd for VGS 0 V?
(b) What is the value of gm0 at VDS 10 V?
ID (mA)
10
VGS = 0 V
9
8
7
6
−1 V
5
4
3
−2 V
2
−3 V
1
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
Figure 9.53 JFET drain characteristic for Problem 12
13. For a 2N4220 n-channel JFET (yfs(minimum) 750 S, yos(maximum) 10 S):
(a) What is the value of gm?
(b) What is the value of rd?
14. (a) Plot gm vs. VGS for an n-channel JFET with IDSS 8 mA and VP 6 V.
(b) Plot gm vs. ID for the same n-channel JFET as part (a).
446
Chapter 9
FET Small-Signal Analysis
−4 V
VDS (V
gm
15. Sketch the ac equivalent model for a JFET if yfs 5.6 mS and yos 15 S.
16. Sketch the ac equivalent model for a JFET if IDSS 10 mA, VP 4 V, VGSQ 2 V, and
yos 25 S.
§ 9.3 JFET Fixed-Bias Configuration
17. Determine Zi, Zo and Av for the network of Fig. 9.54 if IDSS 10 mA, VP 4 V, and rd 40 k.
18. Determine Zi, Zo, and Av for the network of Fig. 9.54 if IDSS 12 mA, VP 6 V, and yos 40 S.
+18 V
1.8 kΩ
Vo
Vi
Zo
1 MΩ
Zi
1.5 V
–
+
Figure 9.54 Fixed-bias amplifier for Problems 17 and 18
§ 9.4 JFET Self-Bias Configuration
19. Determine Zi, Zo, and Av for the network of Fig. 9.55 if yfs 3000 S and yos 50 s.
20. Determine Zi, Zo, and Av for the network of Fig. 9.56 if IDSS 6 mA, VP 6 V, and yos 40 S.
21. Determine Zi, Zo, and Av for the network of Fig. 9.55 if the 20-F capacitor is removed and
the parameters of the network are the same as in Problem 19. Compare results with those of
Problem 19.
22. Repeat Problem 19 if yos is 10 S. Compare the results to those of Problem 19.
+12 V
20 V
2 kΩ
3.3 kΩ
Vo
Vo
Vi
Zi
Zo
Vi
Zo
Zi
10 MΩ
1.1 kΩ
Figure 9.55 Problems 19, 21, and 46
1 MΩ
20 µF
Figure 9.56 Self-bias configuration for
Problems 20 and 47
Problems
447
gm
§ 9.5 JFET Voltage-Divider Configuration
23. Determine Zi, Zo, and Vo for the network of Fig. 9.57 if Vi 20 mV.
+20 V
2 kΩ
82 MΩ
Vo
IDSS = 12 mA
VP = − 3 V
rd = 100 kΩ
Vi
Zi
11 MΩ
RS
610 Ω
Zo
CS
Figure 9.57 Problems 23–26
and 48
24. Determine Zi, Zo, and Vo for the network of Fig. 9.57 if Vi 20 mV and the capacitor CS is
removed.
25. Repeat Problem 23 if rd 20 k and compare results.
26. Repeat Problem 24 if rd 20 k and compare results.
§ 9.6 JFET Source-Follower Configuration
27. Determine Zi, Zo, and Av for the network of Fig. 9.58.
28. Repeat Problem 27 if rd 20 k.
29. Determine Zi, Zo, and Av for the network of Fig. 9.59.
20 V
+20 V
3.3 kΩ
IDSS = 9 mA
VP = −4.5 V
rd = 40 kΩ
Vi
Zi
Vi
Vo
10 MΩ
2.2 kΩ
IDSS = 6 mA
VP = −6 V
rd = 30 kΩ
Zi
3.3 kΩ
Zo
Figure 9.58 Problems 27 and 28
Figure 9.59 Problem 29
§ 9.7 JFET Common-Gate Configuration
30. Determine Zi, Zo, and Vo for the network of Fig. 9.60 if Vi 0.1 mV.
31. Repeat Problem 30 if rd 25 k.
32. Determine Zi, Zo, and Av for the network of Fig. 9.61 if rd 33 k.
448
Chapter 9
FET Small-Signal Analysis
Vo
10 MΩ
Zo
gm
+22 V
+15 V
2.2 kΩ
91 MΩ
3.3 kΩ
Vi
Vi
Zi
Vo
Vo
IDSS = 8 mA
VP = − 2.8 V
rd = 40 kΩ
1.5 kΩ
Zo
Zo
Zi
1 kΩ
11 MΩ
Figure 9.60 Problems 30, 31, and 49
IDSS = 7.5 mA
VP = − 4 V
Figure 9.61 Problem 32
§ 9.8 Depletion-Type MOSFETs
33. Determine Vo for the network of Fig. 9.62 if yos 20 S.
+16 V
1.1 kΩ
Vo
IDSS = 8 mA
VP = − 3 V
+
Vi = 2 mV
10 MΩ
–
Figure 9.62 Problem 33
34. Determine Zi, Zo, and Av for the network of Fig. 9.63 if rd 60 k.
+22 V
1.8 kΩ
Vo
IDSS = 12 mA
VP = − 3.5 V
Vi
Zi
Zo
10 MΩ
100 Ω
Figure 9.63 Problems 34, 35, and 50
Problems
449
gm
35. Repeat Problem 34 if rd 25 k.
36. Determine Vo for the network of Fig. 9.64 if Vi 4 mV.
37. Determine Zi, Zo, and Av for the network of Fig. 9.65.
+18 V
+20 V
6.8 kΩ
91 MΩ
Vo
91 MΩ
yos = 35 µS
yfs = 6000 µS
Vi
Vi
IDSS = 12 mA
VP = −3 V
rd = 45 kΩ
Vo
15 MΩ
Zi
3.3 kΩ
Figure 9.64 Problem 36
10 MΩ
1.1 kΩ
Zo
Figure 9.65 Problem 37
§ 9.10 E-MOSFET Drain-Feedback Configuration
38. Determine gm for a MOSFET if VGS(Th) 3 V and it is biased at VGSQ 8 V. Assume k 0.3 103.
39. Determine Zi, Zo, and Av for the amplifier of Fig. 9.66 if k 0.3 103.
+16 V
2.2 kΩ
Vo
10 MΩ
Zo
VGS(Th) = 3 V
rd = 100 kΩ
Vi
Zi
Figure 9.66 Problems 39, 40, and 51
40. Repeat Problem 39 if k drops to 0.2 103. Compare results.
41. Determine Vo for the network of Fig. 9.67 if Vi 20 mV.
42. Determine Vo for the network of Fig. 9.67 if Vi 4 mV, VGS(Th) 4 V, and ID(on) 4 mA, with
VGS(on) 7 V and yos 20 S.
+20 V
10 kΩ
22 MΩ
Vo
Vi
VGS(Th) = 3.5 V
k = 0.3 × 10−3
yos = 30 µS
Figure 9.67 Problems 41 and 42
450
Chapter 9
FET Small-Signal Analysis
gm
§ 9.11 E-MOSFET Voltage-Divider Configuration
43. Determine the output voltage for the network of Fig. 9.68 if Vi 0.8 mV and rd 40 k.
30V
3.3 kΩ
40 MΩ
+
VGS(Th) = 3 V
k = 0.4 × 10 −3
+
Vo
Vi
10 MΩ
1.2 kΩ
–
–
Figure 9.68 Problem 43
§ 9.12 Designing FET Amplifier Networks
44. Design the fixed-bias network of Fig. 9.69 to have a gain of 8.
45. Design the self-bias network of Fig. 9.70 to have a gain of 10. The device should be biased at
VGSQ 13VP.
+VDD (+22 V)
+VDD (+20 V)
RD
RD
Vo
IDSS = 8 mA
VP = − 2.5 V
yos = 25 µS
Vi
10 MΩ
Vo
IDSS = 12 mA
VP = − 3 V
rd = 40 kΩ
Vi
10 MΩ
RS
Figure 9.69 Problem 44
Figure 9.70 Problem 45
§ 9.13 PSpice Windows
46. Using PSpice Windows, determine the voltage gain for the network of Fig. 9.55.
47. Using PSpice Windows, determine the voltage gain for the network of Fig. 9.56.
48. Using PSpice Windows, determine the voltage gain for the network of Fig. 9.57.
49. Using PSpice Windows, determine the voltage gain for the network of Fig. 9.60.
50. Using PSpice Windows, determine the voltage gain for the network of Fig. 9.63.
51. Using PSpice Windows, determine the voltage gain for the network of Fig. 9.66.
Problems
451
Rs /RL
CHAPTER
10
Systems Approach—
Effects of Rs and RL
10.1 INTRODUCTION
In recent years, the introduction of a wide variety of packaged networks and systems
has generated an increasing interest in the systems approach to design and analysis.
Fundamentally, this approach concentrates on the terminal characteristics of a package and treats each as a building block in the formation of the total package. The content of this chapter is a first step in developing some familiarity with this approach.
The techniques introduced will be used in the remaining chapters and broadened as
the need arises. The trend to packaged systems is quite understandable when you consider the enormous advances in the design and manufacturing of integrated circuits
(ICs). The small IC packages contain stable, reliable, self-testing, sophisticated designs that would be quite bulky if built with discrete (individual) components. The
systems approach is not a difficult one to apply once the basic definitions of the various parameters are correctly understood and the manner in which they are utilized
is clearly demonstrated. In the next few sections, we develop the systems approach
in a slow deliberate manner that will include numerous examples to make each salient
point. If the content of this chapter is clearly and correctly understood, a first plateau
in the understanding of system analysis will be accomplished.
10.2
Ii
Io
+
Zi
AυNL AiNL
Vi
Zo
–
TWO-PORT SYSTEMS
The description to follow can be applied to any two-port system—not only those containing BJTs and FETs—although the emphasis in this chapter is on these active devices. The emphasis in previous chapters on determining the two-port parameters for
various configurations will be quite helpful in the analysis to follow. In fact, many of
the results obtained in the last two chapters are utilized in the analysis to follow.
In Fig. 10.1, the important parameters of a two-port system have been identified.
Note in particular the absence of a load and a source resistance. The impact of these
+ important elements is considered in detail in a later section. For the moment recogVo nize that the impedance levels and the gains of Fig. 10.1 are determined for no-load
(absence of RL) and no-source resistance (Rs) conditions.
–
If we take a “Thévenin look” at the output terminals we find with Vi set to zero
that
Thévenin
Figure 10.1 Two-port system.
452
ZTh Zo Ro
(10.1)
Rs /RL
ETh is the open-circuit voltage between the output terminals identified as Vo. However,
Vo
AvNL Vi
and
so that
Vo AvNLVi
ETh AvNLVi
(10.2)
Note the use of the additional subscript notation “NL” to identify a no-load voltage
gain.
Substituting the Thévenin equivalent circuit between the output terminals will result in the output configuration of Fig. 10.2. For the input circuit the parameters Vi
and Ii are related by Zi Ri, permitting the use of Ri to represent the input circuit.
Since our present interest is in BJT and FET amplifiers, both Zo and Zi can be represented by resistive elements.
Figure 10.2 Substituting the internal elements for the two-port
system of Fig. 10.1.
Before continuing let us check the results of Fig. 10.2 by finding Zo and AvNL in
the usual manner. To find Zo, Vi is set to zero, resulting in AvNLVi 0, permitting
a short-circuit equivalent for the source. The result is an output impedance equal to
Ro as originally defined. The absence of a load will result in Io 0, and the voltage
drop across the impedance Ro will be 0 V. The open-circuit output voltage is therefore
AvNLVi, as it should be. Before looking at an example, take note of the fact that Ai
does not appear in the two-port model of Fig. 10.2 and in fact is seldom part of the
two-port system analysis of active devices. This is not to say that the quantity is seldom calculated, but it is most frequently calculated from the expression Ai Av(Zi/RL), where RL is the defined load for the analysis of interest.
For the fixed-bias transistor network of Fig. 10.3 (Example 8.1), sketch the two-port
equivalent of Fig. 10.2.
EXAMPLE 10.1
Figure 10.3 Example 10.1.
10.2
Two-Port Systems
453
Rs /RL
Solution
From Example 8.1,
Zi 1.069 k
Zo 3 k
AvNL 280.11
Using the information above, the two-port equivalent of Fig. 10.4 can be drawn. Note
in particular the negative sign associated with the controlled voltage source, revealing an opposite polarity for the controlled source than that indicated in the figure. It
also reveals a 180° phase-shift between the input and output voltages.
+
+
Vi
–
Ri
Ro = 3 kΩ
−280.11Vi
1.069 kΩ
–
+
Vo
–
Figure 10.4 Two-port equivalent for the parameters specified
in Example 10.1.
In Example 10.1, RC 3 k was included in defining the no-load voltage gain.
Although this need not be the case (RC could be defined as the load resistor in Chapter 8), the analysis of this chapter will assume that all biasing resistors are part of the
no-load gain and that a loaded system requires an additional load RL connected to the
output terminals.
A second format for Fig. 10.2, particularly popular with op-amps (operational
amplifiers), appears in Fig. 10.5. The only change is the general appearance of the
model.
Figure 10.5 Operational amplifier (op-amp) notation.
10.3 EFFECT OF A LOAD IMPEDANCE (RL)
In this section, the effect of an applied load is investigated using the two-port model
of Fig. 10.2. The model can be applied to any current- or voltage-controlled amplifier. AvNL is, as defined earlier, the gain of the system without an applied load. Ri and
Ro are the input and output impedances of the amplifier as defined by the configuration. Ideally, all the parameters of the model are unaffected by changing loads or
454
Chapter 10
Systems Approach—Effects of Rs and RL
Rs /RL
source resistances (as normally encountered for op-amps to be described in Chapter
14). However, for some transistor amplifier configurations, Ri can be quite sensitive
to the applied load, while for others Ro can be sensitive to the source resistance. In
any case, once AvNL, Ri, and Ro are defined for a particular configuration, the equations about to be derived can be employed.
Applying a load to the two-port system of Fig. 10.2 will result in the configuration of Fig. 10.6. Applying the voltage-divider rule to the output circuit will result in
RLAvNL Vi
Vo RL Ro
and
Vo
RL
Av AvNL
Vi
RL Ro
(10.3)
Figure 10.6 Applying a load to
the two-port system of Fig. 10.2.
Since the ratio RL/(RL Ro) will always be less than 1:
The loaded voltage gain of an amplifier is always less than the no-load level.
Note also that the formula for the voltage gain does not include the input impedance
or current gain.
Although the level of Ri may change with the configuration, the applied voltage
and input current will always be related by
Vi
Vi
Ii Zi
Ri
(10.4)
Defining the output current as the current through the load will result in
Vo
Io RL
(10.5)
with the minus sign occurring due to the defined direction for Io in Fig. 10.6.
The current gain is then determined by
Vo /RL
Io
Vo Zi
Ai Vi/Zi
Ii
Vi RL
and
Zi
Ai Av
RL
(10.6)
for the unloaded situation. In general, therefore, the current gain can be obtained from
the voltage gain and impedance parameters Zi and RL. The next example will demonstrate the usefulness and validity of Eqs. (10.3) through (10.6).
10.3 Effect of a Load Impedance (RL)
455
Rs /RL
EXAMPLE 10.2
In Fig. 10.7, a load has been applied to the fixed-bias transistor amplifier of Example 10.1 (Fig. 10.3).
(a) Determine the voltage and current gain using the two-port systems approach defined by the model of Fig. 10.4.
(b) Determine the voltage and current gain using the re model and compare results.
12 V
RC
RB
Ii
3 kΩ
470 kΩ
Io
+
CC
β = 100
+
Vi
Zo
RL
2.2 kΩ Vo
Zi
–
–
Figure 10.7 Example 10.2.
Solution
(a) Recall from Example 10.1 that
Zi 1.071 k
(with re 10.71 and 100)
Zo 3 k
Av 280.11
NL
Applying Eq. (10.3) yields
RL
Av AvNL
RL Ro
2.2 k
(280.11)
2.2 k 3 k
(0.423)(280.11)
118.5
For the current gain,
Zi
Ai Av RL
In this case, Zi is unaffected by the applied load and
1.071 k
Ai (118.5) 57.69
2.2 k
(b) Substituting the re model will result in the network of Fig. 10.8. Note in particular that the applied load is in parallel with the collector resistor RC defining a
net parallel resistance
RL RC RL 3 k2.2 k 1.269 k
The output voltage
Vo IBRL
456
Chapter 10
Systems Approach—Effects of Rs and RL
Rs /RL
Ib
Ii
+
Vi
Io
RB
+
Zo
Zi
β Ib
β re
470 kΩ
RC
RL
3 kΩ
2.2 kΩ Vo
–
–
R'L
Figure 10.8 Substituting the re model in the ac equivalent network of Fig. 10.7.
with
Vi
Ib re
and
Vi
Vo RL
re
RL
RC RL
Vo
Av re
re
Vi
so that
(10.7)
Substituting values gives
1.269 k
Av 118.5
10.71 as obtained above. For the current gain, by the current-divider rule,
(470 k)Ii
Ib 0.9977Ii Ii
470 k 1.071 k
and
3 k(Ib)
Io 3 k 2.2 k
0.5769Ib
so that
0.5769Ib
0.5769Ii
Io
Ai I
Ii
Ii
i
0.5769(100) 57.69
as obtained using Eq. (10.6).
Example 10.2 demonstrated two techniques to solve the same problem. Although
any network can be solved using the re model approach, the advantage of the systems
approach is that once the two-port parameters of a system are known, the effect of
changing the load can be determined directly from Eq. (10.3). No need to go back to
the ac equivalent model and analyze the entire network. The advantages of the systems approach are similar to those associated with applying Thévenin’s theorem. They
permit concentrating on the effects of the load without having to re-examine the entire network. Of course, if the network of Fig. 10.7 were presented for analysis without the unloaded parameters, it would be a toss-up as to which approach would yield
the desired results in the most direct, efficient manner. However, keep in mind that
the “package” approach is the developing trend. When you purchase a “system” the
two-port parameters are provided, and as with any trend, the user must be aware of
how to utilize the given data.
10.3 Effect of a Load Impedance (RL)
457
Rs /RL
The AC Load Line
For a system such as appearing in Fig. 10.9a, the dc load line was drawn on the output characteristics as shown in Fig. 10.9b. The load resistance did not contribute to
the dc load line since it was isolated from the biasing network by the coupling capacitor (CC). For the ac analysis, the coupling capacitors are replaced by a short-circuit equivalence that will place the load and collector resistors in a parallel arrangement defined by
RL RC RL
The effect on the load line is shown in Fig. 10.9b with the levels to determine the
new axes intersections. Note of particular importance that the ac and dc load lines
pass through the same Q-point—a condition that must be satisfied to ensure a common solution for the network under dc and/or ac conditions.
For the unloaded situation, the application of a relatively small sinusoidal signal
to the base of the transistor could cause the base current to swing from a level of IB2
to IB4 as shown in Fig. 10.9b. The resulting output voltage vce would then have the
swing appearing in the same figure. The application of the same signal for a loaded
situation would result in the same swing in the IB level, as shown in Fig. 10.9b. The
result, however, of the steeper slope of the ac load line is a smaller output voltage
ICQ +
IC
VCEQ
RL'
ac load line
IB
Du
t
et
oV
i
6
IB
5
VCEQ
VCC
RC
IB
iB
R'L
VCC
4
RC
CC
+
3
Q-point
IB
2
CC
RL V
o
Ib
Vi
–
ICQ
dc load line
(a)
0
VCEQ
ICQ R'L
0
vce
vce
t
Figure 10.9 Demonstrating the differences between the dc and ac load lines.
Chapter 10
Systems Approach—Effects of Rs and RL
VCC
VCEQ + ICQ R'L
(b)
458
IB
1
–
0
+
IB
0
RB
VCE
Rs /RL
swing (vce) and a drop in the gain of the system as demonstrated in the numerical
analysis above. It should be obvious from the intersection of the ac load line on the
vertical axis that the smaller the level of RL, the steeper the slope and the smaller the
ac voltage gain. Since RL is smaller for reduced levels of RL, it should be fairly clear
that:
For a particular design, the smaller the level of RL, the lower the level of ac
voltage gain.
10.4 EFFECT OF THE SOURCE IMPEDANCE (RS)
Our attention will now turn to the input side of the two-port system and the effect of
an internal source resistance on the gain of an amplifier. In Fig. 10.10, a source with
an internal resistance has been applied to the basic two-port system. The definitions
of Zi and AvNL are such that:
The parameters Zi and AvNL of a two-port system are unaffected by the internal resistance of the applied source.
Figure 10.10 Including the
effects of the source resistance Rs.
However:
The output impedance may be affected by the magnitude of Rs.
Recall Eq. (8.110) for the complete hybrid equivalent model. The fraction of the
applied signal reaching the input terminals of the amplifier of Fig. 10.10 is determined
by the voltage-divider rule. That is,
RiVs
Vi Ri Rs
(10.8)
Equation (10.8) clearly shows that the larger the magnitude of Rs, the less the voltage at the input terminals of the amplifier. In general, therefore:
For a particular amplifier, the larger the internal resistance of a signal source
the less the overall gain of the system.
For the two-port system of Fig. 10.10,
Vo Av Vi
NL
and
RiVs
Vi Ri Rs
so that
Ri
Vo Av Vs
Ri Rs
and
Vo
Ri
Avs Av
Vs
Ri Rs
NL
NL
(10.9)
10.4 Effect of the Source Impedance (Rs)
459
Rs /RL
The result clearly supports the statement above regarding the reduction in gain with
increase in Rs. Using Eq. (10.9), if Rs 0 (ideal voltage source), Avs Av , which
is the maximum possible value.
The input current is also altered by the presence of a source resistance as follows:
NL
Vs
Ii Rs Ri
EXAMPLE 10.3
(10.10)
In Fig. 10.11, a source with an internal resistance has been applied to the fixed-bias
transistor amplifier of Example 10.1 (Fig. 10.3).
(a) Determine the voltage gain Avs Vo /Vs. What percent of the applied signal appears at the input terminals of the amplifier?
(b) Determine the voltage gain Avs Vo /Vs using the re model.
Figure 10.11 Example 10.3.
Solution
(a) The two-port equivalent for the network appears in Fig. 10.12.
Ro
Rs
+
0.5 kΩ
Vs
+
Vi
–
–
+
Ri
1.071 kΩ
–
3 kΩ
+
AυNLVi
−280.11Vi
Vo
–
Figure 10.12 Substituting the
two-port equivalent network for
the fixed-bias transistor amplifier
of Fig. 10.11.
1.071 k
Ri
Vo
Eq. (10.9): Avs AvNL (280.11)
1.071 k 0.5 k
Vs
Ri Rs
(0.6817)(280.11)
190.96
(1.071 k)Vs
RiVs
Eq. (10.8): Vi 0.6817Vs
1.071 k 0.5 k
Ri Rs
or 68.2% of the available signal reached the amplifier and 31.8% was lost across the
internal resistance of the source.
460
Chapter 10
Systems Approach—Effects of Rs and RL
Rs /RL
(b) Substituting the re model will result in the equivalent circuit of Fig. 10.13. Solving for Vo gives
Vo (100Ib)3 k
Vs
Vs
Zi re and Ib Ii Rs re
1.571 k
with
and
Vs
Vo 100 3 k
1.571 k
so that
(100)(3 k)
Vo
Avs 1.57 k
Vs
190.96
as above.
Ii
Ib
Rs
500 Ω
+
+
Vs
RB
470 kΩ β re
β Ib
100 Ib
1.071 kΩ
–
3 kΩ
Vo
–
Figure 10.13 Substituting the re equivalent circuit for the fixed-bias transistor
amplifier of Fig. 10.11.
Throughout the analysis above, note that Rs was not included in the definition of
Zi for the two-port system. Of course, the resistance “seen” by the source is now Rs
Zi, but Rs remains a quantity associated only with the applied source.
Note again in Example 10.3 that the same results were obtained with the systems
approach and using the re model. Certainly, if the two-port parameters are available,
they should be applied. If not, the approach to the solution is simply a matter of preference.
10.5 COMBINED EFFECT OF Rs AND RL
The effects of Rs and RL have now been demonstrated on an individual basis. The
next natural question is how the presence of both factors in the same network will affect the total gain. In Fig. 10.14, a source with an internal resistance Rs and a load
RL have been applied to a two-port system for which the parameters Zi, Av , and Zo
have been specified. For the moment, let us assume that Zi and Zo are unaffected by
RL and Rs, respectively.
NL
Is
+
Vs
–
Io
Ii
Rs
+
Vi
–
+
Ri
+
Ro
AυNLVi
RL
–
Vo
–
Figure 10.14 Considering the effects of Rs and RL on the gain of an amplifier.
10.5 Combined Effect of Rs and RL
461
Rs /RL
At the input side we find
RiVs
Eq. (10.8): Vi Ri Rs
Vi
Ri
Vs
Ri Rs
or
(10.11)
and at the output side,
RLAv Vi
Vo RL Ro
NL
Vo
RLAV
Av Vi
RL Ro
or
(10.12)
NL
For the total gain Avs Vo /Vs, the following mathematical steps can be performed:
Vo
Vo Vi
Avs Vs
Vi Vs
(10.13)
and substituting Eqs. (10.11) and (10.12) will result in
RLAv
Ri
Avs RL Ro Ri Rs
NL
and
Vo
Ri
RL
Avs Av
Vs
Ri Rs RL Ro
NL
(10.14)
Since Ii Vi /Ri, as before,
Ri
Ai Av RL
(10.15)
Rs Ri
Ais Avs
RL
(10.16)
or using Is Vs /(Rs Ri),
However, Ii Is, so Eqs. (10.15) and (10.16) will generate the same result. Equation
(10.14) clearly reveals that both the source and the load resistance will reduce the
overall gain of the system. In fact:
The larger the source resistance and/or smaller the load resistance, the less
the overall gain of an amplifier.
The two reduction factors of Eq. (10.14) form a product that has to be carefully
considered in any design procedure. It is not sufficient to ensure that Rs is relatively
small if the impact of the magnitude of RL is ignored. For instance, in Eq. (10.14), if
the first factor is 0.9 and the second factor is 0.2, the product of the two results in an
overall reduction factor equal to (0.9)(0.2) 0.18, which is close to the lower factor.
The effect of the excellent 0.9 level was completely wiped out by the significantly
lower second multiplier. If both were 0.9-level factors, the net result would be
(0.9)(0.9) 0.81, which is still quite high. Even if the first were 0.9 and the second
0.7, the net result of 0.63 would still be respectable. In general, therefore, for good
overall gain the effect of both Rs and RL must be evaluated individually and as a
product.
462
Chapter 10
Systems Approach—Effects of Rs and RL
Rs /RL
For the single-stage amplifier of Fig. 10.15, with RL 4.7 k and Rs 0.3 k, determine:
(a) Avs.
(b) Av Vo /Vi.
(c) Ai.
The two-port parameters for the fixed-bias configuration are Zi 1.071 k, Zo 3 k, and AvNL 280.11.
EXAMPLE 10.4
12 V
3 kΩ
470 kΩ
+
0.3 kΩ
Vs
Vo
10 µF
20 µF
β = 100
+
Rs
Zo
Vi
RL
4.7 kΩ
Zi
–
–
Figure 10.15 Example 10.4
Solution
Vo
Ri
RL
(a) Eq. (10.14): Avs Av NL
Vs
Ri Rs RL Ro
1.071 k
4.7 k
(280.11)
1.071 k 0.3 k 4.7 k 3 k
(0.7812)(0.6104)(280.11)
(0.4768)(280.11)
133.57
(4.7 k)(280.11)
Vo
RLAvNL
(b) Av 4.7 k 3 k
Vi
RL Ro
(0.6104)(280.11) 170.98
1.071 k
Ri
(c) Ai Av (170.98) 4.7 k
RL
38.96
1.071 k 0.3 k
Rs Ri
or Ais Avs (133.57) 4.7 k
RL
38.96
as above.
10.6 BJT CE NETWORKS
The fixed-bias configuration has been employed throughout the analysis of the early
sections of this chapter to clearly show the effects of Rs and RL. In this section, various CE configurations are examined with a load and a source resistance. A detailed
analysis will not be performed for each configuration since they follow a very similar path to that demonstrated in the last few sections.
10.6 BJT CE Networks
463
Rs /RL
Fixed Bias
For the fixed-bias configuration examined in detail in recent sections, the system model
with a load and source resistance will appear as shown in Fig. 10.16. In general,
RL
Vo AvNLVi
RL Ro
Rs
Ro
+
Ri
β re
Vi
Aυ NL Vi
Vo
–
Zi
Vs
+
RC
+
Zo
RL
–
+
–
–
Figure 10.16 Fixed-bias
configuration with Rs and RL
Substituting Eq. (8.6), AvNL RC /re and Ro RC,
RL(RC /re)Vi
Vo RL RC
and
Vo
RLRC 1
Av Vi
RL RC re
but
RLRC
RLRC RL RC
and
RLRC
Av r
(10.17)
e
If the re model were substituted for the transistor in the fixed-bias configuration, the
network of Fig. 10.17 would result, clearly revealing that RC and RL are in parallel.
Rs
+
Zi
Vs
Vi
Zo
β re
β Ib
RC
RL
+
Vo
–
+
–
–
RC RL
For the voltage gain Avs of Fig. 10.16,
ZiVs
Vi Zi Rs
464
and
Vi
Zi
Vs
Zi Rs
with
Vo
Vi Vo
Avs Vs
Vs Vi
Chapter 10
Systems Approach—Effects of Rs and RL
Figure 10.17 Fixed-bias
configuration with the
substitution of the re model.
Rs /RL
Zi
Avs Av
Zi Rs
so that
(10.18)
Since the load is connected to the collector terminal of the common-emitter configuration,
and
Zi re
(10.19)
Zo RC
(10.20)
as obtained earlier.
Voltage-Divider Bias
For the loaded voltage-divider bias configuration of Fig. 10.18, the load is again connected to the collector terminal and Zi remains
Zi Rre
(R R1R2)
(10.21)
and for the system’s output impedance
Zo RC
(10.22)
VCC
RC
R1
+
Ib
Rs
C1
+
Vs
–
C2
+
Vi
Zo
R2
Zi
RE
RL Vo
CE
–
–
Figure 10.18 Voltage-divider bias configuration with Rs and RL.
In the small-signal ac model, RC and RL will again be in parallel and
RC RL
Av r
(10.23)
Zi
Avs Av
Zi Rs
(10.24)
e
with
CE Unbypassed Emitter Bias
For the common-emitter unbypassed emitter-bias configuration of Fig. 10.19, Zi remains independent of the applied load and
Zi RBRE
(10.25)
10.6 BJT CE Networks
465
Rs /RL
VCC
RC
RB
Rs
Io
+
Ii
+
C1
+
Vs
Zo
Vi
RL Vo
RE
Zi
–
C2
–
–
Figure 10.19 CE unbypassed emitter-bias configuration with Rs and RL.
For the output impedance,
Zo RC
(10.26)
For the voltage gain, the resistance RC will again drop down in parallel with RL and
RC RL
Vo
Av RE
Vi
(10.27)
with
Vo
Zi
Avs Av
Vs
Zi Rs
(10.28)
and
Io
Zi
Ai Av Ii
RL
(10.29)
but keep in mind that Ii Is Vs /(Rs Zi) Vi/Zi.
Collector Feedback
To keep with our connection of the load to the collector terminal the next configuration to be examined is the collector feedback configuration of Fig. 10.20. In the smallsignal ac model of the system, RC and RL will again drop down in parallel and
VCC
RC
RF
+
C1
Vs
–
C2
Ii
Rs
+
Vi
Zi
Zo
–
Figure 10.20 Collector feedback configuration with Rs and RL.
466
Chapter 10
+
Io
Systems Approach—Effects of Rs and RL
RL
Vo
–
Rs /RL
RC RL
Av r
(10.30)
Zi
Avs Av
Zi Rs
(10.31)
Zo RC RF
(10.32)
RF
Zi re
Av
(10.33)
e
with
The output impedance
and
The fact that Av [Eq. (10.30)] is a function of RL will alter the level of Zi from the
no-load value. Therefore, if the no-load model is available, the level of Zi must be
modified as demonstrated in the next example.
The collector feedback amplifier of Fig. 10.21 has the following no-load system parameters: AvNL 238.94, Zo RCRF 2.66 k, and Zi 0.553 k, with re 11.3 , and 200. Using the systems approach, determine:
(a) Av.
(b) Avs.
(c) Ai.
9V
EXAMPLE 10.5
2.7 kΩ
Io
180 kΩ
Vo
0.6 kΩ
Ii
+
+
Vs
Vi
–
–
β = 200
3.3 kΩ
Figure 10.21 Example 10.5.
Solution
(a) For the two-port system:
2.7 k3.3 k
RC RL
Av 11.3 r
e
1.485 k
131.42
11.3 with
180 k
RF
Zi re (200)(11.3 )
131.42
Av
2.26 k1.37 k
0.853 k
10.6 BJT CE Networks
467
Rs /RL
The system approach will result in the configuration of Fig. 10.22 with the value of
Zi as controlled by RL and the voltage gain. Now the two-port gain equation can be
applied (slight difference in Av due to approximation Ib IRF in Section 8.7):
(3.3 k)(238.94)
RLAvNL
Av 132.3
3.3 k 2.66 k
RL Ro
Ii
+
+
0.6 kΩ
+
+
Vi
Vs
Figure 10.22 The ac equivalent
circuit for the network of Fig.
10.21.
Io
2.66 kΩ
−238.94Vi
0.853 kΩ
–
3.3 kΩ
Vo
–
–
–
0.853 k
Zi
(b) Avs Av (132.3)
0.853
k 0.6 k
Zi Rs
77.67
0.853 k
(132.3)(0.853 k)
Zi
(c) Ai Av (132.3) 3.3
k
3.3 k
RL
34.2
or
Zi Rs
0.853 k 0.6 k
Ai Avs (77.67) RL
3.3 k
34.2
10.7 BJT EMITTER-FOLLOWER
NETWORKS
The input and output impedance parameters of the two-port model for the emitterfollower network are sensitive to the applied load and source resistance. For the emitter-follower configuration of Fig. 10.23, the small-signal ac model would appear as
shown in Fig. 10.24. For the input section of Fig. 10.24, the resistance RB is neglected
because it is usually so much larger than the source resistance that a Thévenin equivVCC
RB
Ii
+
Rs
Vs
Ib
C1
+
Zi
Vi
–
Io
C2
RE
–
Figure 10.23 Emitter-follower
configuraton with Rs and RL.
468
Chapter 10
Systems Approach—Effects of Rs and RL
+
RL Vo
Zo
–
Rs /RL
b
c
Ib
+
β re
Rs
Figure 10.24 Emitter-follower
configuration of Fig. 10.23
following the substitution of
the re equivalent circuit.
β Ib
Vi
+
e
Ie = (β + 1) Ib
Vs
where RE' = RE RL
+
RE' Vo
–
–
–
alent circuit for the configuration of Fig. 10.25 would result in simply Rs and Vs as
shown in Fig. 10.24. Of course, if current levels are to be determined such as Ii in the
original diagram, the effect of RB must be included.
Applying Kirchhoff’s voltage law to the input circuit of Fig. 10.24 will result in
Vs Ib(Rs re ( 1)RE) 0
so that
Vs
Ib Rs re ( 1)RE
+
RB
Vs
Vs IbRs Ibre ( 1)IbRE 0
and
Rs
–
Thévenin
Figure 10.25 Determining the
Thévenin equivalent circuit for
the input circuit of Fig. 10.23.
Establishing Ie, we have
( 1)Vs
Ie ( 1)Ib Rs re ( 1)RE
Vs
Ie [(Rs re)/( 1)] RE
and
Using 1 yields
Vs
Ie (Rs / re) RE
(10.34)
Drawing the network to “fit” Eq. (10.34) will result in the configuration of Fig.
10.26a. In Fig. 10.26b, RE and the load resistance RL have been separated to permit
a definition of Zo and Io.
Rs
β
Rs
+ re
+
β
Ie
R' E
Vs
–
+
Vo
–
+ re
Io
+
Vs
Zo
RE
–
RL
+
Vo
–
(b)
(a)
Figure 10.26 Networks resulting from the application of Kirchhoff’s voltage law to the input circuit of Fig. 10.24.
The voltage gain can then be obtained directly from Fig. 10.26a using the voltage
divider rule:
REVs
Vo RE (Rs / re)
or
RE
Vo
Avs RE (Rs / re)
Vs
10.7 BJT Emitter-Follower Networks
469
Rs /RL
RERL
Vo
Avs R
R
Vs
E
L Rs/ re
and
(10.35)
Setting Vs 0 and solving for Zo will result in
Rs
Zo RE re
(10.36)
For the input impedance,
Zb (re RE)
Zi RBZb
and
Zi RB(re RE RL)
or
(10.37)
For no-load conditions, the gain equation is
RE
AvNL RE re
while for loaded conditions,
RE RL
Vo
Av RE RL re
Vi
EXAMPLE 10.6
(10.38)
For the loaded emitter-follower configuration of Fig. 10.27 with a source resistance
and the no-load two-port parameters of Zi 157.54 k, Zo 21.6 , and AvNL 0.993 with re 21.74 and 65, determine:
(a) The new values of Zi and Zo as determined by the load and Rs, respectively.
(b) Av using the systems approach.
(c) Avs using the systems approach.
15 V
(d) Ai Io /Ii.
560 kΩ
Ii
0.56 kΩ
C1
β = 65
C2
+
+
Vs
–
Vo
Zi
Io
Vi
3.3 kΩ
–
Figure 10.27 Example 10.6.
Zo
Solution
Eq. (10.37):
Zi RB(re RE RL)







560 k65(21.74 3.3 k2.2 k)
560 k87.21 k
75.46 k
versus 157.54 k (no-load).
470
Chapter 10
Systems Approach—Effects of Rs and RL
1.32 k
2.2 kΩ
Rs /RL
Rs
Zo RE re
0.56 k
3.3 k 21.74 65
3.3 k30.36 30.08 versus 21.6 (no Rs).
(b) Substituting the two-port equivalent network will result in the small-signal ac
equivalent network of Fig. 10.28.
(2.2 k)(0.993)Vi
RLAvNLVi
Vo 2.2 k 30.08 RL Ro
0.98Vi
Vo
with Av 0.98
Vi
Ii
+
Vs
–
Io
30.08 Ω
0.56 kΩ
+
+
+
Vi
0.993 Vi
75.46 kΩ
–
–
Vo
2.2 kΩ
–
Figure 10.28 Small-signal ac equivalent circuit for the network of Fig. 10.27.
(75.46 k)Vs
ZiVs
(c) Vi 0.993Vs
75.46 k 0.56 k
Zi Rs
Vo
Vo Vi
Avs (0.98)(0.993) 0.973
Vs
Vi Vs
so that
Io
Zi
(d) Ai Av Ii
RL
75.46 k
(0.98) 2.2 k
33.61
10.8 BJT CB NETWORKS
A common-base amplifier with an applied load and source resistance appear in Fig.
10.29. The fact that the load is connected between the collector and base terminals
isolates it from the input circuit, and Zi remains essentially the same for no-load or
loaded conditions. The isolation that exists between input and output circuits also
maintains Zo at a fixed level even though the level of Rs may change. The voltage
gain is now determined by
10.8 BJT CB Networks
471
Rs /RL
Ii
+
Rs
C1
Vs
+
RE
Vi
Zi
–
RC
VEE
–
+
Io
C2
RL
Zo
VCC
Vo
–
Figure 10.29 Common-base configuration with Rs and RL.
RCRL
Av re
(10.39)
Ai 1
(10.40)
and the current gain:
EXAMPLE 10.7
For the common-base amplifier of Fig. 10.30, the no-load two-port parameters are
(using 1) Zi re 20 , AvNL 250, and Zo 5 k. Using the two-port equivalent model, determine:
(a) Av.
(b) Avs.
(c) Ai.
Ii
+
0.2 kΩ 10 µF
Vs
Io
+
α≈
=1
1 kΩ
Vi
–
+
10 µF
5 kΩ
2V
–
Vo
8.2 kΩ
8V
–
Figure 10.30 Example 10.7.
Solution
(a) The small-signal ac equivalent network appears in Fig. 10.31.
(8.2 k)(250)Vi
RLAvNLVi
Vo 8.2 k 5 k 155.3Vi
RL Ro
Ii
+
0.2 kΩ
Vs
–
Io
+
Vi
Zi
–
+
1 kΩ
20 Ω
+
5 kΩ
250Vi
8.2 kΩ
–
Figure 10.31 Small-signal ac equivalent circuit for the network of Fig. 10.30.
472
Chapter 10
Systems Approach—Effects of Rs and RL
Vo
–
Rs /RL
Vo
Av 155.3
Vi
5 k8.2 k
RCRL
3.106 k
Av 20
re
20 and
or
155.3
Vo
Vi Vo
(b) Avs Vs
Vs Vi
20 Ri
Av (155.3)
20 200 Ri Rs
14.12
Note the relatively low gain due to a source impedance much larger than the input
impedance of the amplifier.
20 Zi
(c) Ai Av (155.3) 8.2
k
RL
0.379
which is significantly less than 1 due to the division of output current between RC
and RL.
10.9 FET NETWORKS
As noted in Chapter 9, the isolation that exists between gate and drain or source of
an FET amplifier ensures that changes in RL do not affect the level of Zi and changes
in Rsig do not affect Ro. In essence, therefore:
The no-load two-port model of Fig. 10.2 for an FET amplifier is unaffected by
an applied load or source resistance.
Bypassed Source Resistance
For the FET amplifier of Fig. 10.32, the applied load will appear in parallel with RD
in the small-signal model, resulting in the following equation for the loaded gain:
Av gm(RDRL)
(10.41)
VDD
RD
+
C2
C1
Vs
Zi
Vi
Zo
RG
RS
RL V
o
CS
–
Rsig
+
+
–
–
Figure 10.32 JFET amplifier with Rsig and RL.
10.9 FET Networks
473
Rs /RL
The impedance levels remain at
Zi RG
(10.42)
Zo RD
(10.43)
Unbypassed Source Resistance
For the FET amplifier of Fig. 10.33, the load will again appear in parallel with RD
and the loaded gain becomes
gm(RDRL)
Vo
Av 1 gmRS
Vi
(10.44)
with
Zi RG
(10.45)
and
Zo RD
(10.46)
VDD
RD
Rsig
Vs
Zi
Zo
RG
Vi
RL
RS
–
–
Figure 10.33 JFET amplifier
with unbypassed RS.
Vo
–
C1
+
+
EXAMPLE 10.8
+
C2
For the FET amplifier of Fig. 10.34, the no-load two-port parameters are AvNL 3.18, Zi R1R2 239 k, and Zo 2.4 k, with gm 2.2 mS.
(a) Using the two-port parameters above, determine Av and Avs.
(b) Using Eq. (10.44), calculate the loaded gain and compare to the result of part (a).
VDD
R1
10 µF
+
Zo
RS
Zi
1
Vi
R2
–
RS
2
RL
Chapter 10
4.7 kΩ Vo
CS = 20 µ F
–
474
0.3 kΩ
270 kΩ
1.2 kΩ
Figure 10.34 Example 10.8.
+
20 µ F
Systems Approach—Effects of Rs and RL
–
1 kΩ
Vs
2.4 kΩ
+
Rsig
RD
2.1 MΩ
Rs /RL
Solution
(a) The small-signal ac equivalent network appears in Fig. 10.35, and
(4.7 k)(3.18)
RLAvNL
Vo
Av 4.7 k 2.4 k
Vi
RL Ro
2.105
Vo
Vi Vo
Ri
Avs Av
Vs
Vs Vi
Ri Rsig
(239 k)(2.105)
239 k 1 k
2.096 Av
1 kΩ
+
+
Zo
− 3.18Vi
239 kΩ
Vi
2.4 kΩ
4.7 kΩ Vo
–
Vs
+
Zi
–
+
–
–
Figure 10.35 Small-signal ac equivalent circuit for the network of Fig. 10.34.
gm(RDRL)
(b) Eq. (10.44): Av 1g R
m S1
(2.2 mS)(2.4 k4.7 k)
3.498
1 (2.2 mS)(0.3 k)
1.66
2.105 as above
Source Follower
For the source-follower configuration of Fig. 10.36, the level of Zi is independent of
the magnitude of RL and determined by
Zi RG
(10.47)
VCC
Rsig
+
C1
Vs
–
RG
Zi
+
C2
RS
Zo
RL
Vo
–
Figure 10.36 Source-follower configuration with Rsig and RL.
10.9 FET Networks
475
Rs /RL
The loaded voltage gain has the same format as the unloaded gain with RS replaced
by the parallel combination of RS and RL.
gm(RS RL)
Vo
Av 1
gm(RS RL)
Vi
(10.48)
The level of output impedance is as determined in Chapter 9:
1
Zo RS gm
(10.49)
revealing an insensitivity to the magnitude of the source resistance Rsig.
Common Gate
Even though the common-gate configuration of Fig. 10.37 is somewhat different from
those described above with regard to the placement of RL and Rsig, the input and output circuits remain isolated and
RS
Zi 1 gmRS
(10.50)
Zo RD
(10.51)
Av gm(RDRL)
(10.52)
The loaded voltage gain is given by
VDD
RD
+
C2
Rsig
+
Vs
–
C1
Zi
Zo
+
Vi
RL
Vo
RS
–
–
Figure 10.37 Common-gate configuration with Rsig and RL.
10.10 SUMMARY TABLE
Now that the loaded and unloaded (Chapters 8 and 9) BJT and JFET amplifiers have
been examined in some detail, a review of the equations developed is provided by
Table 10.1. Although all the equations are for the loaded situation, the removal of RL
will result in the equations for the unloaded amplifier. The same is true for the effect
of Rs (for BJTs) and Rsig (for JFETs) on Zo. In each case, the phase relationship between the input and output voltages is also provided for quick reference. A review of
476
Chapter 10
Systems Approach—Effects of Rs and RL
Rs /RL
the equations will reveal that the isolation provided by the JFET between the gate and
channel by the SiO2 layer results in a series of less complex equations than those encountered for the BJT configurations. The linkage provided by Ib between input and
output circuits of the BJT transistor amplifier adds a touch of complexity to some of
the equations.
TABLE 10.1 Summary of Transistor Configurations (Av, Zi, Zo)
Configuration
Av Vo /Vi
Zi
Zo
RLRC)
re
RBre
RC
hfe
(RLRC)
hie
RBhie
RC
(RLRCro)
r
RBre
RC ro
(RLRC)
re
R1R2re
RC
hfe
(RLRC)
hie
R1R2hie
RC
R1R2re
RC ro
R
E RLRE
Rs RsR1R2
1
R1R2(re R
E)
Rs
RE re
1
R1R2(hie hfeR
E)
Rs hie
RE hfe
1
R1R2(re R
E)
Rs
RE re
(RLRC)
r
RE re
RC
hfb
h (RLRC)
RE hib
RC
Including ro:
(RLRCro)
re
RE re
RC ro
Including ro:
e
Including ro:
(RLRCro)
re
Including ro:
e
ib
10.10 Summary Table
477
Rs /RL
TABLE 10.1 Summary of Transistor Configurations (Av, Zi, Zo) (Continued)
Configuration
VCC
Av Vo /Vi
Zi
Zo
RLRC)
RE
R1R2(re RE)
RC
(RLRC)
RE
R1R2(hie hfeRE)
RC
R1R2(re RE)
RC
RLRC
RE1
RB(re RE1)
RC
(RLRC)
RE1
RB(hie hfeRE1)
RC
RB(re RE1)
RC
(RLRC)
re
RF
re Av
RC
hfe
(RLRC)
hie
RF
hie
A RC
Including ro:
(RLRCro)
re
RF
re Av
RC RF ro
(RLRC)
RE
RF
RE Av
RC RF
(RLRC)
RE
RF
hfeRE Av
RC RF
RF
RE Av
RC RF
RC
R1
Rs
Vo
Vi
Zo
+
RL
Zi
Vs
R2
RE
–
Including ro:
(RLRC)
RE
VCC
RC
RB
Rs
Vo
Vi
Zo
+
Zi
Vs
–
RL
RE1
RE2
Including ro:
RLRC
RE1
CE
VCC
RC
RF
Rs
Vo
Vi
Zo
+
Vs
RL
Zi
–
v
VCC
RC
RF
Rs
Vo
Vi
Zo
+
Vs
–
478
RL
Zi
RE
L
Including ro:
(RLRC)
RE
Chapter 10
Systems Approach—Effects of Rs and RL
Rs /RL
TABLE 10.1 (Continued)
Configuration
Av Vo /Vi
Zi
Zo
gm(RDRL)
RG
RD
gm(RDRLrd)
RG
RDrd
gm(RDRL)
1 gmRS
RG
RD
1 gmRS
gm(RDRL)
RD RS
1 gmRS rd
RG
RD
1 gmRS
gm(RDRL)
R1R2
RD
gm(RDRLrd)
R1R2
RDrd
gm(RSRL)
1 gm(RSRL)
RG
RS1/gm
Including rd:
gmrd (RSRL)
rd RD gmrd (RS RL)
RG
RS
gmrdRS
1 rd RD
RS
1 gmRS
RD
VDD
RD
Vo
Rsig
Vi
Zo
+
RL
Vs
Zi
–
RG
Including rd:
RS
CS
VDD
RD
Vo
Rsig
Vi
Zo
+
RL
Vs
Zi
–
RG
RS
Including rd:
VDD
RD
R1
Rsig
Vo
Vi
Zo
+
RL
Vs
Zi
–
Including rd:
R2
RS
CS
VDD
RD
Rsig
Vi
+
Vs
Zi
–
Rsig
Vo
RG
RS
Vi
Vo
+
Vs
–
RL
Zo
Zi
RD
RS
VDD
Zo
RL
gm(RDRL)
Including rd:
gm(RDRL)
RS
Zi gmrdRS
1 rd RDRL
10.10 Summary Table
RDrd
479
Rs /RL
10.11 CASCADED SYSTEMS
The two-port systems approach is particularly useful for cascaded systems such as
that appearing in Fig. 10.38, where Av1, Av2, Av3, and so on, are the voltage gains of
each stage under loaded conditions. That is, Av1 is determined with the input impedance to Av2 acting as the load on Av1. For Av2, Av1 will determine the signal strength
and source impedance at the input to Av2. The total gain of the system is then determined by the product of the individual gains as follows:
AvT Av1 Av2 Av3
(10.53)
Zi1
AiT AvT
RL
(10.54)
and the total current gain by
No matter how perfect the system design, the application of a load to a two-port
system will affect the voltage gain. Therefore, there is no possibility of a situation
where Av1, Av2, and so on, of Fig. 10.38 are simply the no-load values. The loading
of each succeeding stage must be considered. The no-load parameters can be used to
determine the loaded gains of Fig. 10.38, but Eq. (10.53) requires the loaded values.
Vo = Vi
1
Vo = Vi
2
2
3
+
+
Aυ
Vi
Aυ
1
Aυ
2
A υn
3
RL Vo
–
–
Zi = Zi
1
Zo
1
Zi
2
Zo
2
Zi
3
Zo
3
Z in
Zon = Zo
Figure 10.38 Cascaded system.
EXAMPLE 10.9
The two-stage system of Fig. 10.39 employed a transistor emitter-follower configuration prior to a common-base configuration to ensure that the maximum percent of
the applied signal appears at the input terminals of the common-base amplifier. In
Fig. 10.39, the no-load values are provided for each system, with the exception of Zi
and Zo for the emitter-follower, which are the loaded values. For the configuration of
Fig. 10.39, determine:
(a) The loaded gain for each stage.
(b) The total gain for the system, Av and Avs.
(c) The total current gain for the system.
(d) The total gain for the system if the emitter-follower configuration were removed.
Figure 10.39 Example 10.9.
480
Chapter 10
Systems Approach—Effects of Rs and RL
Rs /RL
Solution
(a) For the emitter-follower configuration, the loaded gain is
(26 )(1)Vi1
Zi2 AvNLVi1
Vo1 26
12 0.684Vi1
Zi2 Zo1
Vo1
Av 0.684
Vi1
and
1
For the common-base configuration,
(8.2 k)(240)Vi2
RLAvNLVi2
Vo2 8.2 k 5.1 k 147.97Vi2
RL Ro2
Vo2
AvL 147.97
Vi2
and
(b) AvT Av1Av2
(0.684)(147.97)
101.20
(10 k)(101.20)
Zi1
Avs AvT 10 k 1 k
Zi1 Rs
92
10 k
Zi1
(c) AiT AvT
(101.20) 8.2
k
RL
123.41
(26 )Vs
ZiCBVs
(d) ViCB 0.025Vs
26 1 k
ZiCB Rs
Vi
and 0.025
Vs
with
Vo
147.97
Vi
from above
Vi Vo
and Avs (0.025)(147.97) 3.7
Vs Vi
In total, therefore, the gain is about 25 times greater with the emitter-follower configuration to draw the signal to the amplifier stages. Take note, however, that it was
also important that the output impedance of the first stage was relatively close to the
input impedance of the second stage or the signal would have been “lost” again by
the voltage-divider action.
10.12 PSPICE WINDOWS
Loaded Voltage-Divider BJT Transistor Configuration
The computer analysis of this section includes a PSpice Windows evaluation of the
response of a loaded BJT and FET amplifier with a source resistance. The BJT network of Fig. 10.40 employs the same unloaded configuration examined in the PSpice
analysis of Chapter 8, where the unloaded gain was 369 (Example 8.2, re 18.44 ).
For the transistor, all the parameters listed under Model Editor were removed except
Is and beta, which were set to 2E-15A and 90, respectively. In this way, the results
10.12 PSpice Windows
481
Rs /RL
will be as close to the hand-written solutions as possible without going to the controlled source equivalents. Note the placement of the VPRINT1 option to pick up the
voltage lost across the source resistance and to note if there is any drop in gain across
the capacitor. The option Do not auto-run Probe was chosen, and under Analysis
Setup, the AC Sweep was set at a fixed frequency of 10 kHz. In addition, Display
Results on Schematic under Analysis was chosen and the Voltage Display enabled.
Figure 10.40 Loaded
voltage-divider BJT transistor
configuration.
An Analysis resulted in the dc levels appearing in Fig. 10.40. In particular, note
the zero volt levels at the left side of C1 and the right side of C2. In addition, note
that VBE is essentially 0.7 V and the dc levels of each terminal of the transistor are
very close to those calculated in Example 8.2 (using the approximate approach). Reviewing the output file following Analysis-Examine Output will result in the data
listings of Fig. 10.41. The nodes are defined in the Schematics Netlist, and the BJT
MODEL PARAMETERS reveal our choices for this run—although the last three
are default values. The SMALL-SIGNAL BIAS SOLUTION simply confirms the
levels printed on the schematic, and the Operating Point Information reveals that
beta (dc and ac) is 90, that VBE is 0.7 V, that IC is 1.32 mA, and that IB is 14.7 A
(in addition to a host of other levels). The AC ANALYSIS reveals that the voltage
on the other side of Rsig is about 0.7 mV, resulting in a drop of about 0.3 mV (30%
loss in signal voltage) of the applied signal across Rsig. The remaining two ac levels
are the same, revealing that the capacitor is an effective short circuit for ac. The loaded
gain from source to output is 144.9. The gain from the base of the transistor to the
output is 144.9 mV/0.7 mV 207. Both levels are certainly significantly less than
the no-load level of 369. If we return to the network and change RL to 10 M, the
output voltage will rise to 243.3 mV, resulting in a gain of 243.3 mV/0.7 mV 347.57,
which is quite close to the hand-calculated, approximate level of 369.
For interest’s sake, let us now calculate the loaded voltage gain and compare to
the PSpice solution of 144.9.
re 18.44 and
Zi R1R2re
56 k8.2 k(90)(18.44 )
1.35 k
(1.35 k)Vs
ZiVs
Vi 1.35
k 0.6 k 0.69Vs
Zi Rs
and
482
Chapter 10
Vi
0.69
Vs
Systems Approach—Effects of Rs and RL
Rs /RL
Figure 10.41 Output file for
the network of Fig. 10.40.
483
Rs /RL
(10 k)(350.4)
Vo
RLAvNL
Av 10 k 6.8 k
Vi
RL Ro
208.57
with
Vi Vo
Avs (0.69)(208.57)
Vs Vi
144
which is an excellent comparison with the computer solution.
Loaded JFET Self-Bias Transistor Configuration
The network of Fig. 10.42 is a loaded version of the network examined in Chapter 9,
which resulted in a no-load gain of 13.3. In the Model Editor dialog box, Beta was
set to 0.625mA/V2 and Vto 4V. The remaining parameters were left alone to permit a close comparison with the Chapter 9 solution and because they have less effect
on the response than for a BJT transistor.
Again, note the effectiveness of the capacitors to block the dc voltages. In addition, note the small voltage at the gate, indicating that the input impedance to the de-
Figure 10.42 Loaded self-bias
JFET transistor configuration.
vice is in reality not infinite (although for all practical purposes it is an excellent assumption.) Again, the frequency was set to 10 kHz and an Analysis called for without the Probe option. The sequence Analysis-Examine Output will result in the listing of Fig. 10.43. The Schematics Netlist provides a listing of assigned nodes, and
the OPERATING POINT INFORMATION reveals that the drain current is 3.36
mA, that VGS is 1.71 V, and that gm is 2.94 mS. The AC ANALYSIS reveals that
there is negligible drop across either capacitor at this frequency, and the short-circuit
equivalency can be assumed. The output voltage is 5.597mV resulting in a loaded
gain of 5.597 compared to the unloaded gain of 13.3. Note also that the drop across
Rsig is negligible due to the high input impedance of the device.
Using the value of gm hand-calculated earlier, the equation for the loaded gain
will result in a gain of 5.62 as shown below—an excellent comparison with the computer solution.
Av gm(RDRL)
(2.90 mS)(4.7 k3.3 k)
5.62
The results obtained above have clearly substantiated the analysis and equations
presented in this chapter for a loaded amplifier.
484
Chapter 10
Systems Approach—Effects of Rs and RL
Rs /RL
Figure 10.43 Output file for
the network of Fig. 10.42.
Rs /RL
PROBLEMS
§ 10.3 Effect of a Load Impedance (RL)
1. For the fixed-bias configuration of Fig. 10.44:
(a) Determine AvNL, Zi, and Zo.
(b) Sketch the two-port model of Fig. 10.2 with the parameters determined in part (a) in place.
(c) Calculate the gain Av using the model of part (b) and Eq. (10.3).
(d) Determine the current gain using Eq. (10.6).
(e) Determine Av, Zi, and Zo using the re model and compare with the solutions above.
18 V
3.3 kΩ
680 kΩ
1.8 µF
Vi
Ii
1.8 µF
Io
Vo
β = 100
Zo
RL
4.7 kΩ
Zi
Figure 10.44 Problems 1, 2, and 3
* 2. (a) Draw the dc and ac load lines for the network of Fig. 10.44 on the characteristics of Fig.
10.45.
(b) Determine the peak-to-peak value of Ic and Vce from the graph if Vi has a peak value of
10 mV. Determine the voltage gain Av Vo/Vi and compare with the solution obtained in
Problem 1.
Figure 10.45 Problems 2 and 7
486
Chapter 10
Systems Approach—Effects of Rs and RL
Rs /RL
3. (a) Determine the voltage gain Av for the network of Fig. 10.44 for RL 4.7, 2.2, and 0.5 k.
What is the effect of decreasing levels of RL on the voltage gain?
(b) How will Zi, Zo, and AvNL change with decreasing values of RL?
§ 10.4 Effect of a Source Impedance (Rs)
* 4. For
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
the network of Fig. 10.46:
Determine AvNL, Zi, and Zo.
Sketch the two-port model of Fig. 10.2 with the parameters determined in part (a) in place.
Determine Av using the results of part (b).
Determine Avs.
Determine Avs using the re model and compare the results to that obtained in part (d).
Change Rs to 1 k and determine Av. How does Av change with the level of Rs?
Change Rs to 1 k and determine Avs. How does Avs change with the level of Rs?
Change Rs to 1 k and determine AvNL, Zi, and Zo. How do they change with change in Rs?
12 V
3 kΩ
1 MΩ
Rs
1 µF
1 µF
Vo
Vi
β = 180
0.6 kΩ
+
Zo
Vs
Zi
–
Figure 10.46 Problem 4
§ 10.5 Combined Effect of Rs and RL
* 5. For the network of Fig. 10.47:
(a) Determine AvNL, Zi, and Zo.
(b) Sketch the two-port model of Fig. 10.2 with the parameters determined in part (a) in place.
(c) Determine Av and Avs.
(d) Calculate Ai.
(e) Change RL to 5.6 k and calculate Avs. What is the effect of increasing levels of RL on the
gain?
(f) Change Rs to 0.5 k (with RL at 2.7 k) and comment on the effect of reducing Rs on Avs.
(g) Change RL to 5.6 k and Rs to 0.5 k and determine the new levels of Zi and Zo. How are
the impedance parameters affected by changing levels of RL and Rs?
24 V
4.3 kΩ
560 kΩ
Ii
+
Rs
10 µF
Io
Vo
Vi
β = 80
1 kΩ
Zo
Vs
–
10 µ F
RL
2.7 kΩ
Zi
Figure 10.47 Problems 5 and 17
Problems
487
Rs /RL
§ 10.6 BJT CE Networks
6. For the voltage-divider configuration of Fig. 10.48:
(a) Determine AvNL, Zi, and Zo.
(b) Sketch the two-port model of Fig. 10.2 with the parameters determined in part (a) in place.
(c) Calculate the gain Av using the model of part (b).
(d) Determine the current gain Ai.
(e) Determine Av, Zi, and Zo using the re model and compare solutions.
Figure 10.48 Problems 6, 7, and 8
* 7. (a) Draw the dc and ac load lines for the network of Fig. 10.48 on the characteristics of Fig.
10.45.
(b) Determine the peak-to-peak value of Ic and Vce from the graph if Vi has a peak value of 10
mV. Determine the voltage gain Av Vo /Vi and compare the solution with that obtained in
Problem 6.
8. (a) Determine the voltage gain Av for the network of Fig. 10.48 with RL 4.7, 2.2, and 0.5
k. What is the effect of decreasing levels of RL on the voltage gain?
(b) How will Zi, Zo, and AvNL change with decreasing levels of RL?
9. For the emitter-stabilized network of Fig. 10.49:
(a) Determine AvNL, Zi, and Zo.
(b) Sketch the two-port model of Fig. 10.2 with the values determined in part (a).
(c) Determine Av and Avs.
(d) Change Rs to 1 k. What is the effect on AvNL, Zi, and Zo?
(e) Change Rs to 1 k and determine Av and Avs. What is the effect of increasing levels of Rs
on Av and Avs?
Figure 10.49 Problem 9
488
Chapter 10
Systems Approach—Effects of Rs and RL
Rs /RL
§ 10.7 BJT Emitter-Follower Networks
* 10. For the network of Fig. 10.50:
(a) Determine AvNL, Zi, and Zo.
(b) Sketch the two-port model of Fig. 10.2 with the values determined in part (a).
(c) Determine Av and Avs.
(d) Change Rs to 1 k and determine Av and Avs. What is the effect of increasing levels of Rs
on the voltage gains?
(e) Change Rs to 1 k and determine AvNL, Zi, and Zo. What is the effect of increasing levels
of Rs on the parameters?
(f) Change RL to 5.6 k and determine Av and Avs. What is the effect of increasing levels of
RL on the voltage gains? Maintain Rs at its original level of 0.6 k.
Figure 10.50 Problems 10 and 18
§ 10.8 BJT CB Networks
* 11. For the common-base network of Fig. 10.51:
(a) Determine Zi, Zo, and AvNL.
(b) Sketch the two-port model of Fig. 10.2 with the parameters of part (a) in place.
(c) Determine Av and Avs.
(d) Determine Av and Avs using the re model and compare with the results of part (c).
(e) Change Rs to 0.5 k and RL to 2.2 k and calculate Av and Avs. What is the effect of changing levels of Rs and RL on the voltage gains?
(f) Determine Zo if Rs changed to 0.5 k with all other parameters as appearing in Fig. 10.51.
How is Zo affected by changing levels of Rs?
(g) Determine Zi if RL is reduced to 2.2 k. What is the effect of changing levels of RL on the
input impedance?
Figure 10.51 Problems 11 and 19
Problems
489
Rs /RL
§ 10.9 FET Networks
12. For the self-bias JFET network of Fig. 10.52:
(a) Determine AvNL, Zi, and Zo.
(b) Sketch the two-port model of Fig. 10.2 with the parameters determined in part (a) in place.
(c) Determine Av and Avs.
(d) Change RL to 6.8 k and Rsig to 1 k and calculate the new levels of Av and Avs. How are
the voltage gains affected by changes in Rsig and RL?
(e) For the same changes as part (d), determine Zi and Zo. What was the impact on both impedances?
12 V
2.7 kΩ
10 µ F
Vo
Rsig
Vi
0.6 kΩ
IDSS = 10 mA
VP = − 6 V
10 µF
+
Zo
Vs
1 MΩ
Zi
0.51 kΩ
RL
4.7 kΩ
20 µF
–
Figure 10.52 Problems 12 and 20
13. For the source-follower network of Fig. 10.53:
(a) Determine AvNL, Zi, and Zo.
(b) Sketch the two-port model of Fig. 10.2 with the parameters determined in part (a) in place.
(c) Determine Av and Avs.
(d) Change RL to 4.7 k and calculate Av and Avs. What was the effect of increasing levels of
RL on both voltage gains?
(e) Change Rsig to 1 k (with RL at 2.2 k) and calculate Av and Avs. What was the effect of
increasing levels of Rsig on both voltage gains?
(f) Change RL to 4.7 k and Rsig to 1 k and calculate Zi and Zo. What was the effect on both
parameters?
12 V
8.2 µF
Rsig
+
IDSS = 6 mA
VP = − 6 V
Vi
0.5 kΩ
Vo
Vs
2 MΩ
Zi
8.2 µF
3.3 kΩ
Zo
–
Figure 10.53 Problem 13
490
Chapter 10
Systems Approach—Effects of Rs and RL
2.2 kΩ
Rs /RL
* 14. For the common-gate configuration of Fig. 10.54:
(a) Determine AvNL, Zi, and Zo.
(b) Sketch the two-port model of Fig. 10.2 with the parameters determined in part (a) in place.
(c) Determine Av and Avs.
(d) Change RL to 2.2 k and calculate Av and Avs. What was the effect of changing RL on the
voltage gains?
(e) Change Rsig to 0.5 k (with RL at 4.7 k) and calculate Av and Avs. What was the effect
of changing Rsig on the voltage gains?
(f) Change RL to 2.2 k and Rsig to 0.5 k and calculate Zi and Zo. What was the effect on
both parameters?
18 V
3.3 kΩ
5.6 µF
Vo
IDSS = 5 mA
VP = − 4 V
5.6 µ F
Rsig
+
Zo
Vi
4.7 kΩ
1 kΩ
Vs
1.2 kΩ
Zi
–
Figure 10.54 Problem 14
§ 10.11 Cascaded Systems
* 15. For the cascaded system of Fig. 10.55 with two identical stages, determine:
(a) The loaded voltage gain of each stage.
(b) The total gain of the system, Av and Avs.
(c) The loaded current gain of each stage.
(d) The total current gain of the system.
(e) How Zi is affected by the second stage and RL.
(f) How Zo is affected by the first stage and Rs.
(g) The phase relationship between Vo and Vi.
Rs
+
1 µF V
i
1 µF
CE amplifier
0.6 kΩ
Vs
Zi
Vo
CE amplifier
Zi = 1 kΩ
Zi = 1 kΩ
Zo = 3.3 kΩ
Zo = 3.3 kΩ
Aυ NL = – 420
Aυ NL = – 420
–
Zo
1
Zo
RL
2.7 kΩ
Zi
2
Figure 10.55 Problem 15
Problems
491
Rs /RL
* 16. For the cascaded system of Fig. 10.56, determine:
(a) The loaded voltage gain of each stage.
(b) The total gain of the system, Av and Avs.
(c) The loaded current gain of each stage.
(d) The total current gain of the system.
(e) How Zi is affected by the second stage and RL.
(f) How Zo is affected by the first stage and Rs.
(g) The phase relationship between Vo and Vi.
Rs
+
10 µF
Vi
Emitter - follower
1 kΩ 10 µF
Vs
Zi
Vo
CE amplifier
Zi = 50 kΩ
Z i = 1.2 kΩ
Zo = 20 Ω
Aυ ≅ 1
Zo = 4.6 kΩ
Aυ NL = – 640
–
NL
Zo
1
RL
2.2 kΩ
Zo
Zi
2
Figure 10.56 Problem 16
§ 10.12 PSpice Windows
17. Using PSpice Windows, determine the level of Vo for Vs 1 mV for the network of Fig. 10.47.
For the capacitive elements assume a frequency of 1 kHz.
18. Repeat Problem 17 for the network of Fig. 10.50 and compare the results with those of Problem 10.
19. Repeat Problem 17 for the network of Fig. 10.51 and compare with the results of Problem 11.
20. Repeat Problem 17 for the network of Fig. 10.52 and compare with the results of Problem 12.
*Please Note: Asterisks indicate more difficult problems.
492
Chapter 10
Systems Approach—Effects of Rs and RL
f
CHAPTER
BJT and JFET
Frequency Response
11
11.1 INTRODUCTION
The analysis thus far has been limited to a particular frequency. For the amplifier, it
was a frequency that normally permitted ignoring the effects of the capacitive elements, reducing the analysis to one that included only resistive elements and sources
of the independent and controlled variety. We will now investigate the frequency
effects introduced by the larger capacitive elements of the network at low frequencies and the smaller capacitive elements of the active device at the high frequencies.
Since the analysis will extend through a wide frequency range, the logarithmic scale
will be defined and used throughout the analysis. In addition, since industry typically
uses a decibel scale on its frequency plots, the concept of the decibel is introduced
in some detail. The similarities between the frequency response analyses of both BJTs
and FETs permit a coverage of each in the same chapter.
11.2 LOGARITHMS
There is no escaping the need to become comfortable with the logarithmic function.
The plotting of a variable between wide limits, comparing levels without unwieldy
numbers, and identifying levels of particular importance in the design, review, and
analysis procedures are all positive features of using the logarithmic function.
As a first step in clarifying the relationship between the variables of a logarithmic function, consider the following mathematical equations:
a bx,
x logb a
(11.1)
The variables a, b, and x are the same in each equation. If a is determined by taking the base b to the x power, the same x will result if the log of a is taken to the base
b. For instance, if b 10 and x 2,
a bx (10)2 100
but
x logb a log10 100 2
In other words, if you were asked to find the power of a number that would result in
a particular level such as shown below:
10,000 10x
493
f
the level of x could be determined using logarithms. That is,
x log10 10,000 4
For the electrical/electronics industry and in fact for the vast majority of scientific research, the base in the logarithmic equation is limited to 10 and the number e 2.71828. . . .
Logarithms taken to the base 10 are referred to as common logarithms, while
logarithms taken to the base e are referred to as natural logarithms. In summary:
Common logarithm: x log10 a
(11.2)
Natural logarithm: y loge a
(11.3)
loge a 2.3 log10 a
(11.4)
The two are related by
On today’s scientific calculators, the common logarithm is typically denoted by the
key and the natural logarithm by the
key.
EXAMPLE 11.1
Using the calculator, determine the logarithm of the following numbers to the base
indicated.
(a) log10 106.
(b) loge e3.
(c) log10 102.
(d) loge e1.
Solution
(a) 6
(b) 3
(c) 2
(d) 1
The results in Example 11.1 clearly reveal that the logarithm of a number taken
to a power is simply the power of the number if the number matches the base of the
logarithm. In the next example, the base and the variable x are not related by an integer power of the base.
EXAMPLE 11.2
Using the calculator, determine the logarithm of the following numbers.
(a) log10 64.
(b) loge 64.
(c) log10 1600.
(d) log10 8000.
Solution
(a) 1.806
(b) 4.159
(c) 3.204
(d) 3.903
Note in parts (a) and (b) of Example 11.2 that the logarithms log10 a and loge a
are indeed related as defined by Eq. (11.4). In addition, note that the logarithm of a
number does not increase in the same linear fashion as the number. That is, 8000 is
125 times larger than 64, but the logarithm of 8000 is only about 2.16 times larger
494
Chapter 11
BJT and JFET Frequency Response
f
than the magnitude of the logarithm of 64, revealing a very nonlinear relationship. In
fact, Table 11.1 clearly shows how the logarithm of a number increases only as the
exponent of the number. If the antilogarithm of a number is desired, the 10x or ex calculator functions are employed.
TABLE 11.1
log10 100
log10 10
log10 100
log10 1,000
log10 10,000
log10 100,000
log10 1,000,000
log10 10,000,000
log10 100,000,000
and so on
0
1
2
3
4
5
6
7
8
EXAMPLE 11.3
Using a calculator, determine the antilogarithm of the following expressions:
(a) 1.6 log10 a.
(b) 0.04 loge a.
Solution
(a) a 101.6
Calculator keys:
and a 39.81
(b) a e0.04
Calculator keys:
and a 1.0408
Since the remaining analysis of this chapter employs the common logarithm, let
us now review a few properties of logarithms using solely the common logarithm. In
general, however, the same relationships hold true for logarithms to any base.
log10 1 0
(11.5)
As clearly revealed by Table 11.1, since 100 1,
a
log10 log10 a log10 b
b
(11.6)
which for the special case of a 1 becomes
1
log10 log10 b
b
(11.7)
revealing that for any b greater than 1 the logarithm of a number less than 1 is always negative.
log10 ab log10 a log10 b
(11.8)
In each case, the equations employing natural logarithms will have the same format.
11.2 Logarithms
495
f
EXAMPLE 11.4
Using a calculator, determine the logarithm of the following numbers:
(a) log10 0.5.
4000
(b) log10 .
250
(c) log10 (0.6 30).
Solution
(a) 0.3
(b) log10 4000 log10 250 3.602 2.398 1.204
4000
Check: log10 log10 16 1.204
250
(c) log10 0.6 log10 30 0.2218 1.477 1.255
Check: log10 (0.6 30) log10 18 1.255
The use of log scales can significantly expand the range of variation of a particular variable on a graph. Most graph paper available is of the semilog or double-log
(log-log) variety. The term semi (meaning one-half) indicates that only one of the two
scales is a log scale, whereas double-log indicates that both scales are log scales. A
semilog scale appears in Fig. 11.1. Note that the vertical scale is a linear scale with
equal divisions. The spacing between the lines of the log plot is shown on the graph.
Figure 11.1 Semilog graph paper.
496
Chapter 11 BJT and JFET Frequency Response
The log of 2 to the base 10 is approximately 0.3. The distance from 1 (log10 1 0)
to 2 is therefore 30% of the span. The log of 3 to the base 10 is 0.4771 or almost
48% of the span (very close to one-half the distance between power of 10 increments
on the log scale). Since log10 5 0.7, it is marked off at a point 70% of the distance.
Note that between any two digits the same compression of the lines appears as you
progress from the left to the right. It is important to note the resulting numerical value
and the spacing, since plots will typically only have the tic marks indicated in Fig.
11.2 due to a lack of space. You must realize that the longer bars for this figure have
the numerical values of 0.3, 3, and 30 associated with them, whereas the next shorter
bars have values of 0.5, 5, and 50 and the shortest bars 0.7, 7, and 70.
(3)
about halfway (0.3)
0.1
0.7
(5) (7)
(30) (50) (70)
10
1
f
100
log
almost three-fourths (0.5)
Figure 11.2
Identifying the numerical values of the tic marks on a log scale.
Be aware that plotting a function on a log scale can change the general appearance of the waveform as compared to a plot on a linear scale. A straight-line plot on
a linear scale can develop a curve on a log scale, and a nonlinear plot on a linear scale
can take on the appearance of a straight line on a log plot. The important point is that
the results extracted at each level be correctly labeled by developing a familiarity with
the spacing of Figs. 11.1 and 11.2. This is particularly true for some of the log-log
plots that appear later in the book.
11.3 DECIBELS
The concept of the decibel (dB) and the associated calculations will become increasingly important in the remaining sections of this chapter. The background surrounding the term decibel has its origin in the established fact that power and audio levels
are related on a logarithmic basis. That is, an increase in power level, say 4 to 16 W,
does not result in an audio level increase by a factor of 16/4 4. It will increase by
a factor of 2 as derived from the power of 4 in the following manner: (4)2 16. For
a change of 4 to 64 W, the audio level will increase by a factor of 3 since (4)3 64.
In logarithmic form, the relationship can be written as log4 64 3.
The term bel was derived from the surname of Alexander Graham Bell. For standardization, the bel (B) was defined by the following equation to relate power levels
P1 and P2:
P2
G log10 P1
bel
(11.9)
11.3 Decibels
497
f
It was found, however, that the bel was too large a unit of measurement for practical purposes, so the decibel (dB) was defined such that 10 decibels 1 bel.
Therefore,
P2
GdB 10 log10 P1
dB
(11.10)
The terminal rating of electronic communication equipment (amplifiers, microphones, etc.) is commonly rated in decibels. Equation (11.10) indicates clearly, however, that the decibel rating is a measure of the difference in magnitude between two
power levels. For a specified terminal (output) power (P2) there must be a reference
power level (P1). The reference level is generally accepted to be 1 mW, although on
occasion, the 6-mW standard of earlier years is applied. The resistance to be associated with the 1-mW power level is 600 , chosen because it is the characteristic impedance of audio transmission lines. When the 1-mW level is employed as the reference level, the decibel symbol frequently appears as dBm. In equation form,
P2
GdBm 10 log10 1 mW 600 dBm
(11.11)
There exists a second equation for decibels that is applied frequently. It can be
best described through the system of Fig. 11.3. For Vi equal to some value V1, P1 V 12/Ri, where Ri , is the input resistance of the system of Fig. 11.3. If Vi should be increased (or decreased) to some other level, V2, then P2 V22/Ri. If we substitute into
Eq. (11.10) to determine the resulting difference in decibels between the power
levels,
P2
V 22/Ri
V2
GdB 10 log10 10 log10 2 10 log10 P1
V1
V 1/Ri
and
V2
GdB 20 log10 V1
dB
2
(11.12)
Figure 11.3 Configuration employed in the discussion of Eq. (11.12).
Frequently, the effect of different impedances (R1 R2) is ignored and Eq. (11.12)
applied simply to establish a basis of comparison between levels—voltage or current.
For situations of this type, the decibel gain should more correctly be referred to as
the voltage or current gain in decibels to differentiate it from the common usage of
decibel as applied to power levels.
One of the advantages of the logarithmic relationship is the manner in which it
can be applied to cascaded stages. For example, the magnitude of the overall voltage
gain of a cascaded system is given by
(11.13)
Av Av Av Av …Av T
498
Chapter 11
1
2
BJT and JFET Frequency Response
3
n
f
Applying the proper logarithmic relationship results in
TABLE 11.2
Gv 20 log10 AvT 20 log10 Av1 20 log10 Av2
20 log10 Av3 Voltage Gain,
20 log10 Avn
(dB)
(11.14)
In words, the equation states that the decibel gain of a cascaded system is simply the
sum of the decibel gains of each stage, that is,
Gv Gv1 Gv2 Gv3 Gvn
dB
(11.15)
In an effort to develop some association between dB levels and voltage gains,
Table 11.2 was developed. First note that a gain of 2 results in a dB level of 6 dB
while a drop to 12 results in a 6-dB level. A change in Vo /Vi from 1 to 10, 10 to 100,
or 100 to 1000 results in the same 20-dB change in level. When Vo Vi, Vo /Vi 1
and the dB level is 0. At a very high gain of 1000, the dB level is 60, while at the
much higher gain of 10,000, the dB level is 80 dB, an increase of only 20 dB—a result of the logarithmic relationship. Table 11.2 clearly reveals that voltage gains of
50 dB or higher should immediately be recognized as being quite high.
Vo /Vi
dB Level
0.5
0.707
1
2
10
40
100
1000
10,000
etc.
6
3
0
6
20
32
40
60
80
EXAMPLE 11.5
Find the magnitude gain corresponding to a decibel gain of 100.
Solution
By Eq. (11.10),
P2
P2
GdB 10 log10 100 dB → log10 10
P1
P1
so that
P2
1010 10,000,000,000
P1
This example clearly demonstrates the range of decibel values to be expected from
practical devices. Certainly, a future calculation giving a decibel result in the neighborhood of 100 should be questioned immediately.
The input power to a device is 10,000 W at a voltage of 1000 V. The output power
is 500 W, while the output impedance is 20 .
(a) Find the power gain in decibels.
(b) Find the voltage gain in decibels.
(c) Explain why parts (a) and (b) agree or disagree.
EXAMPLE 11.6
Solution
Po
500 W
1
(a) GdB 10 log10 10 log10 10 log10 10 log10 20
Pi
10 kW
20
10(1.301) 13.01 dB
Vo
P
R
(5
00
W2
)(0)
(b) Gv 20 log10 20 log10 20 log10 Vi
1000
1000 V
100
1
20 log10 20 log10 20 log10 10 20 dB
1000
10
Vi2
(1 kV)2
106
(c) Ri 4 100 Ro 20 Pi
10 kW
10
11.3 Decibels
499
f
EXAMPLE 11.7
An amplifier rated at 40-W output is connected to a 10- speaker.
(a) Calculate the input power required for full power output if the power gain is
25 dB.
(b) Calculate the input voltage for rated output if the amplifier voltage gain is 40 dB.
Solution
(a) Eq. (11.10):
(b) Gv 20 log10
40 W
25 10 log10 Pi
40 W
40 W
Pi antilog (2.5)
3.16 102
40 W
126.5 mW
316
Vo
Vo
40 20 log10 Vi
Vi
Vo
antilog 2 100
Vi
Vo P
R
(4
0)(
W10)
V 20 V
Vo
20 V
Vi 0.2 V 200 mV
100
100
11.4 GENERAL FREQUENCY
CONSIDERATIONS
The frequency of the applied signal can have a pronounced effect on the response of
a single-stage or multistage network. The analysis thus far has been for the midfrequency spectrum. At low frequencies, we shall find that the coupling and bypass capacitors can no longer be replaced by the short-circuit approximation because of the
increase in reactance of these elements. The frequency-dependent parameters of the
small-signal equivalent circuits and the stray capacitive elements associated with the
active device and the network will limit the high-frequency response of the system.
An increase in the number of stages of a cascaded system will also limit both the
high- and low-frequency responses.
The magnitudes of the gain response curves of an RC-coupled, direct-coupled,
and transformer-coupled amplifier system are provided in Fig. 11.4. Note that the horizontal scale is a logarithmic scale to permit a plot extending from the low- to the
high-frequency regions. For each plot, a low-, high-, and mid-frequency region has
been defined. In addition, the primary reasons for the drop in gain at low and high
frequencies have also been indicated within the parentheses. For the RC-coupled amplifier, the drop at low frequencies is due to the increasing reactance of CC, Cs, or CE,
while its upper frequency limit is determined by either the parasitic capacitive elements of the network and frequency dependence of the gain of the active device. An
explanation of the drop in gain for the transformer-coupled system requires a basic
understanding of “transformer action” and the transformer equivalent circuit. For the
moment, let us say that it is simply due to the “shorting effect” (across the input terminals of the transformer) of the magnetizing inductive reactance at low frequencies
(XL 2 fL). The gain must obviously be zero at f 0 since at this point there is no
longer a changing flux established through the core to induce a secondary or output
voltage. As indicated in Fig. 11.4, the high-frequency response is controlled primarily by the stray capacitance between the turns of the primary and secondary wind500
Chapter 11
BJT and JFET Frequency Response
f
Figure 11.4 Gain versus frequency: (a) RC-coupled amplifiers; (b) transformercoupled amplifiers; (c) direct-coupled amplifiers.
ings. For the direct-coupled amplifier, there are no coupling or bypass capacitors to
cause a drop in gain at low frequencies. As the figure indicates, it is a flat response
to the upper cutoff frequency, which is determined by either the parasitic capacitances
of the circuit or the frequency dependence of the gain of the active device.
For each system of Fig. 11.4, there is a band of frequencies in which the magnitude of the gain is either equal or relatively close to the midband value. To fix the
frequency boundaries of relatively high gain, 0.707Avmid was chosen to be the gain at
the cutoff levels. The corresponding frequencies f1 and f2 are generally called the corner, cutoff, band, break, or half-power frequencies. The multiplier 0.707 was chosen
because at this level the output power is half the midband power output, that is, at
midfrequencies,
AvmidVi2
V2o
Pomid Ro
Ro
and at the half-power frequencies,
0.707AvmidVi2
AvmidVi2
PoHPF 0.5 Ro
Ro
11.4 General Frequency Considerations
501
f
PoHPF 0.5Pomid
and
(11.16)
The bandwidth (or passband) of each system is determined by f1 and f2, that is,
bandwidth (BW) f2 f1
(11.17)
For applications of a communications nature (audio, video), a decibel plot of the
voltage gain versus frequency is more useful than that appearing in Fig. 11.4. Before
obtaining the logarithmic plot, however, the curve is generally normalized as shown
in Fig. 11.5. In this figure, the gain at each frequency is divided by the midband value.
Obviously, the midband value is then 1 as indicated. At the half-power frequencies,
the resulting level is 0.707 1/2. A decibel plot can now be obtained by applying
Eq. (11.12) in the following manner:
Av
Av
dB 20 log10 Avmid
Avmid
(11.18)
Aυ
A υmid
1
0.707
f1 100
10
1000
10,000
100,000
f2
1 MHz
10 MHz
f (log scale)
Figure 11.5 Normalized gain versus frequency plot.
At midband frequencies, 20 log10 1 0, and at the cutoff frequencies, 20 log10
1/2 3 dB. Both values are clearly indicated in the resulting decibel plot of Fig.
11.6. The smaller the fraction ratio, the more negative the decibel level.
Aυ
Aυ
10
0 dB
mid (dB)
f1
100
1000
10,000
100,000
f2
1 MHz
10 MHz
f (log scale)
− 3 dB
− 6 dB
− 9 dB
− 12 dB
Figure 11.6 Decibel plot of the normalized gain versus frequency plot of Fig. 11.5.
For the greater part of the discussion to follow, a decibel plot will be made only
for the low- and high-frequency regions. Keep Fig. 11.6 in mind, therefore, to permit
a visualization of the broad system response.
It should be understood that most amplifiers introduce a 180° phase shift between
input and output signals. This fact must now be expanded to indicate that this is the
case only in the midband region. At low frequencies, there is a phase shift such that
Vo lags Vi by an increased angle. At high frequencies, the phase shift will drop below 180°. Figure 11.7 is a standard phase plot for an RC-coupled amplifier.
502
Chapter 11
BJT and JFET Frequency Response
f
< (Vo leads Vi )
360°
270°
180°
90°
0°
10
1000
f1 100
10,000
100,000
f2
1 MHz
10 MHz
f (log scale)
Figure 11.7 Phase plot for an RC-coupled amplifier system.
11.5 LOW-FREQUENCY ANALYSIS—
BODE PLOT
In the low-frequency region of the single-stage BJT or FET amplifier, it is the R-C
combinations formed by the network capacitors CC, CE, and Cs and the network resistive parameters that determine the cutoff frequencies. In fact, an R-C network similar to Fig. 11.8 can be established for each capacitive element and the frequency at
which the output voltage drops to 0.707 of its maximum value determined. Once the
cutoff frequencies due to each capacitor are determined, they can be compared to establish which will determine the low-cutoff frequency for the system.
Our analysis, therefore, will begin with the series R-C combination of Fig. 11.8
and the development of a procedure that will result in a plot of the frequency response
with a minimum of time and effort. At very high frequencies,
Figure 11.8 R-C combination
that will define a low cutoff frequency.
+
Vi
–
1
XC 0 2 fC
and the short-circuit equivalent can be substituted for the capacitor as shown in Fig.
11.9. The result is that Vo Vi at high frequencies. At f 0 Hz,
+
R
Vo
–
Figure 11.9 R-C circuit of Figure 11.8 at very high frequencies.
1
1
XC 2 fC
2 (0)C
and the open-circuit approximation can be applied as shown in Fig. 11.10, with the
result that Vo 0 V.
Between the two extremes, the ratio Av Vo /Vi will vary as shown in Fig. 11.11.
As the frequency increases, the capacitive reactance decreases and more of the input
voltage appears across the output terminals.
Figure 11.10 R-C circuit of
Figure 11.8 at f 0 Hz.
A υ = Vo / Vi
1
0.707
0
f1
f
Figure 11.11 Low frequency response for the R-C circuit of Figure 11.8.
11.5 Low-Frequency Analysis—Bode Plot
503
f
The output and input voltages are related by the voltage-divider rule in the following manner:
RVi
Vo R XC
with the magnitude of Vo determined by
RVi
Vo 2
R
2
X
C
For the special case where XC R,
RVi
RVi
RVi
RVi
1
Vi
Vo 2 2
2
R
2R
X C
2
R 2
Vo
1
Av 0.707XCR
Vi
2
and
(11.19)
the level of which is indicated on Fig. 11.11. In other words, at the frequency of which
XC R, the output will be 70.7% of the input for the network of Fig. 11.8.
The frequency at which this occurs is determined from
1
XC R
2 f1C
1
f1 2 RC
and
(11.20)
In terms of logs,
1
Gv 20 log10 Av 20 log10 3 dB
2
while at Av Vo /Vi 1 or Vo Vi (the maximum value),
Gv 20 log10 1 20(0) 0 dB
In Fig. 11.6, we recognize that there is a 3-dB drop in gain from the midband
level when f f1. In a moment, we will find that an RC network will determine the
low-frequency cutoff frequency for a BJT transistor and f1 will be determined by
Eq. (11.20).
If the gain equation is written as
Vo
R
1
1
1
Av Vi
R jXC
1 j(XC/R)
1 j(1/ CR)
1 j(1/2 fCR)
and using the frequency defined above,
1
Av 1 j( f1/f )
(11.21)
Vo
1
Av 2 tan1( f1/f )
Vi
1
(
f1/f
)
(11.22)
magnitude of Av
504
Chapter 11
BJT and JFET Frequency Response












In the magnitude and phase form,
phase by which
Vo leads Vi
f
For the magnitude when f f1,
1
1
Av 2 0.707 → 3 dB
1
2
(1
)
In the logarithmic form, the gain in dB is
f1
1
Av(dB) 20 log10 2 20 log10 1 f
1
( f1/f
)
2 1/2
f1
(12)(20) log10 1 f
2
f1
10 log10 1 f
For frequencies where f
by
f1 or (f1/f)2
2
1, the equation above can be approximated
f1
Av(dB) 10 log10 f
and finally,
2
f1
Av(dB) 20 log10 f
(11.23)
f
f1
Ignoring the condition f f1 for a moment, a plot of Eq. (11.23) on a frequency
log scale will yield a result of a very useful nature for future decibel plots.
f1
At f f1: 1 and 20 log10 1 0 dB
f
f1
At f 12 f1: 2 and 20 log10 2 6 dB
f
f1
At f 14 f1: 4 and 20 log10 4 12 dB
f
f1
At f 110 f1: 10 and 20 log10 10 20 dB
f
A plot of these points is indicated in Fig. 11.12 from 0.1f1 to f1. Note that this results in a straight line when plotted against a log scale. In the same figure, a straight
Figure 11.12 Bode plot for the
low-frequency region.
505
f
line is also drawn for the condition of 0 dB for f f1. As stated earlier, the straightline segments (asymptotes) are only accurate for 0 dB when f f1 and the sloped line
when f1 f. We know, however, that when f f1, there is a 3-dB drop from the midband level. Employing this information in association with the straight-line segments
permits a fairly accurate plot of the frequency response as indicated in the same figure. The piecewise linear plot of the asymptotes and associated breakpoints is called
a Bode plot of the magnitude versus frequency.
The calculations above and the curve itself demonstrate clearly that:
A change in frequency by a factor of 2, equivalent to 1 octave, results in a
6-dB change in the ratio as noted by the change in gain from f1/2 to f1.
As noted by the change in gain from f1/2 to f1:
For a 10:1 change in frequency, equivalent to 1 decade, there is a 20-dB
change in the ratio as demonstrated between the frequencies of f1/10 and f1.
In the future, therefore, a decibel plot can easily be obtained for a function having the format of Eq. (11.23). First, simply find f1 from the circuit parameters and
then sketch two asymptotes—one along the 0-dB line and the other drawn through f1
sloped at 6 dB/octave or 20 dB/decade. Then, find the 3-dB point corresponding to
f1 and sketch the curve.
EXAMPLE 11.8
For the network of Fig. 11.13:
(a) Determine the break frequency.
(b) Sketch the asymptotes and locate the 3-dB point.
(c) Sketch the frequency response curve.
Solution
Figure 11.13 Example 11.8
1
1
(a) f1 3
2 RC
(6.28)(5 10 )(0.1 106 F)
318.5 Hz
(b) and (c). See Fig. 11.14.
Figure 11.14 Frequency response for the R-C circuit of Figure 11.13.
506
Chapter 11
BJT and JFET Frequency Response
f
The gain at any frequency can then be determined from the frequency plot in the
following manner:
Vo
Av(dB) 20 log10 Vi
but
Av(dB)
Vo
log10 Vi
20
and
Vo
20
Av 10
Vi
Av(dB)
(11.24)
For example, if Av(dB) 3 dB,
Vo
Av 10(3/20) 10(0.15) 0.707
Vi
as expected
The quantity 100.15 is determined using the 10x function found on most scientific
calculators.
From Fig. 11.14, Av(dB) 1 dB at f 2f1 637 Hz. The gain at this point is
Vo
20
10(1/20) 10(0.05) 0.891
Av 10
Vi
Av(dB)
Vo 0.891Vi
and
or Vo is 89.1% of Vi at f 637 Hz.
The phase angle of is determined from
f1
tan1 f
from Eq. (11.22).
For frequencies f
(11.25)
f1,
f1
tan1 → 90°
f
For instance, if f1 100f,
f1
tan1 tan1(100) 89.4°
f
For f f1,
f1
tan1 tan11 45°
f
For f
f1,
f1
tan1 → 0°
f
For instance, if f 100f1,
f1
tan1 tan1 0.01 0.573°
f
11.5 Low-Frequency Analysis—Bode Plot
507
f
A plot of tan1(f1/f) is provided in Fig. 11.15. If we add the additional 180°
phase shift introduced by an amplifier, the phase plot of Fig. 11.7 will be obtained.
The magnitude and phase response for an R-C combination have now been established. In Section 11.6, each capacitor of importance in the low-frequency region will
be redrawn in an R-C format and the cutoff frequency for each determined to establish the low-frequency response for the BJT amplifier.
Figure 11.15 Phase response for the R-C circuit of Figure 11.8.
11.6 LOW-FREQUENCY RESPONSE —
BJT AMPLIFIER
The analysis of this section will employ the loaded voltage-divider BJT bias configuration, but the results can be applied to any BJT configuration. It will simply be necessary to find the appropriate equivalent resistance for the R-C combination. For the
network of Fig. 11.16, the capacitors Cs, CC, and CE will determine the low-frequency
response. We will now examine the impact of each independently in the order listed.
VCC
RC
R1
CC
Vo
Cs
+
Rs
RL
Vi
+
Vs
–
R2
RE
Zi
CE
–
Figure 11.16 Loaded BJT amplifier with capacitors that affect the low-frequency response.
Cs
Since Cs is normally connected between the applied source and the active device, the
general form of the R-C configuration is established by the network of Fig. 11.17.
The total resistance is now Rs Ri, and the cutoff frequency as established in Section 11.5 is
508
Chapter 11
BJT and JFET Frequency Response
1
fLS 2 (Rs Ri)Cs
f
(11.26)
At mid or high frequencies, the reactance of the capacitor will be sufficiently small +
to permit a short-circuit approximation for the element. The voltage Vi will then be Vs
related to Vs by
–
RiVs
Vimid Ri Rs
(11.27)
At fLS, the voltage Vi will be 70.7% of the value determined by Eq. (11.27), assuming that Cs is the only capacitive element controlling the low-frequency response.
For the network of Fig. 11.16, when we analyze the effects of Cs we must make
the assumption that CE and CC are performing their designed function or the analysis becomes too unwieldy, that is, that the magnitude of the reactances of CE and CC
permits employing a short-circuit equivalent in comparison to the magnitude of the
other series impedances. Using this hypothesis, the ac equivalent network for the input section of Fig. 11.16 will appear as shown in Fig. 11.18.
The value of Ri for Eq. (11.26) is determined by
Cs
+
Cs
Rs
Vi
Ri
System
–
Figure 11.17 Determining the
effect of Cs on the low frequency
response.
+
Rs
+
Vi
R2 R2
hie = βre
Vs
Figure 11.18 Localized ac equivalent for Cs.
–
–
Ri R1R2re
(11.28)
The voltage Vi applied to the input of the active device can be calculated using the
voltage-divider rule:
RiVs
Vi Rs Ri jXCS
(11.29)
+
CC
RL Vo
System
CC
Ro
Since the coupling capacitor is normally connected between the output of the active
device and the applied load, the R-C configuration that determines the low cutoff frequency due to CC appears in Fig. 11.19. From Fig. 11.19, the total series resistance
is now Ro RL and the cutoff frequency due to CC is determined by
Figure 11.19 Determining the
effect of CC on the low-frequency
response.
(11.30)
(11.31)
C
+
ro
Vc
–
Ignoring the effects of Cs and CE, the output voltage Vo will be 70.7% of its midband
value at fLC. For the network of Fig. 11.16, the ac equivalent network for the output
section with Vi 0 V appears in Fig. 11.20. The resulting value for Ro in Eq. (11.30)
is then simply
Ro RC ro
Thévenin
+
CC
RL
RC
Ro
Figure 11.20 Localized ac
equivalent for CC with Vi 0 V.
11.6 Low-Frequency Response—BJT Amplifier
Vo
–
1
fLC 2 (Ro RL)CC
–
509
f
CE
System
CE
Re
1
fLE 2 ReCE
Figure 11.21 Determining the
effect of CE on the low-frequency
response.
R's
+ re
β
To determine fLE , the network “seen” by CE must be determined as shown in Fig.
11.21. Once the level of Re is established, the cutoff frequency due to CE can be determined using the following equation:
For the network of Fig. 11.16, the ac equivalent as “seen” by CE appears in Fig. 11.22.
The value of Re is therefore determined by
Rs
Re RE re
E
Re
RE
CE
RC
Av re RE
RC
Vo
Vi
RE
Figure 11.23 Network employed
to describe the effect of CE on
the amplifier gain.
The maximum gain is obviously available where RE is zero ohms. At low frequencies, with the bypass capacitor CE in its “open-circuit” equivalent state, all of RE appears in the gain equation above, resulting in the minimum gain. As the frequency increases, the reactance of the capacitor CE will decrease, reducing the parallel
impedance of RE and CE until the resistor RE is effectively “shorted out” by CE. The
result is a maximum or midband gain determined by Av RC/re. At fLE the gain
will be 3 dB below the midband value determined with RE “shorted out.”
Before continuing, keep in mind that Cs, CC, and CE will affect only the lowfrequency response. At the midband frequency level, the short-circuit equivalents for
the capacitors can be inserted. Although each will affect the gain Av Vo/Vi in a similar frequency range, the highest low-frequency cutoff determined by Cs, CC, or CE
will have the greatest impact since it will be the last encountered before the midband
level. If the frequencies are relatively far apart, the highest cutoff frequency will essentially determine the lower cutoff frequency for the entire system. If there are two
or more “high” cutoff frequencies, the effect will be to raise the lower cutoff frequency and reduce the resulting bandwidth of the system. In other words, there is an
interaction between capacitive elements that can affect the resulting low cutoff frequency. However, if the cutoff frequencies established by each capacitor are sufficiently separated, the effect of one on the other can be ignored with a high degree of
accuracy—a fact that will be demonstrated by the printouts to appear in the following example.
(a) Determine the lower cutoff frequency for the network of Fig. 11.16 using the following parameters:
Cs 10 F,
Rs 1 k,
RL 2.2 k
100,
CE 20 F,
R1 40 k,
ro ,
CC 1 F
R2 10 k,
VCC 20 V
(b) Sketch the frequency response using a Bode plot.
510
(11.33)
where Rs RsR1R2.
The effect of CE on the gain is best described in a quantitative manner by recalling that the gain for the configuration of Fig. 11.23 is given by
Figure 11.22 Localized ac
equivalent of CE.
EXAMPLE 11.9
(11.32)
Chapter 11
BJT and JFET Frequency Response
RE 2 k,
RC 4 k,
f
Solution
(a) Determining re for dc conditions:
RE (100)(2 k) 200 k
10R2 100 k
The result is:
R2VCC
10 k(20 V)
200 V
VB 4 V
R2 R1
10 k 40 k
50
VE
4 V 0.7 V
3.3 V
IE 1.65 mA
2 k
RE
2 k
with
26 mV
re 15.76 1.65 mA
so that
re 100(15.76 ) 1576 1.576 k
and
Midband Gain
Vo
RCRL
(4 k) (2.2 k)
Av 90
Vi
re
15.76 The input impedance
Zi Ri R1R2re
40 k10 k1.576 k
1.32 k
and from Fig. 11.24,
RiVs
Vi Ri Rs
Vi
Ri
1.32 k
0.569
1.32 k 1 k
Vs
Ri Rs
or
Vo
Vo Vi
Avs (90)(0.569)
Vs
Vi Vs
so that
51.21
Rs
+
+
Vs
Ri
Vi
–
–
Figure 11.24 Determining the
effect of Rs on the gain Avs.
Cs
Ri R1R2re 40 k10 k1.576 k 1.32 k
1
1
fLS 2 (Rs Ri)Cs
(6.28)(1 k 1.32 k)(10 F)
fLS 6.86 Hz
11.6 Low-Frequency Response—BJT Amplifier
511
f
The results just obtained will now be verified using PSpice Windows. The network with its various capacitors appears in Fig. 11.25. The Model Editor was used
to set Is to 2E-15A and beta to 100. The remaining parameters were removed from
the listing to idealize the response to the degree possible. Under Analysis Setup-AC
Sweep, the frequency was set to 10 kHz to establish a frequency in the midband region. A simulation of the network resulted in the dc levels of Fig. 11.25. Note that
VB is 3.9 V versus the calculated level of 4 V and that VE is 3.2 V versus the calculated level of 3.3 V. Very close when you consider that the approximate model was
used. VBE is very close to the 0.7 V at 0.71 V. The output file reveals that the ac voltage across the load at a frequency of 10 kHz is 49.67 mV, resulting in a gain of 49.67,
which is very close to the calculated level of 51.21.
Figure 11.25 Network of Figure 11.16 with assigned values.
A plot of the gain versus frequency will now be obtained with only CS as a determining factor. The other capacitors, CC and CE, will be set to very high values so
they are essentially short circuits at any of the frequencies of interest. Setting CC and
CE to 1 F will remove any affect they will have on the response on the low-frequency
region. Here, one must be careful as the program does not recognize 1F as one Farad.
It must be entered as 1E6uF. Since the pattern desired is gain versus frequency, we
must use the sequence Analysis-Setup-Analysis Setup-Enable AC Sweep-AC Sweep
to obtain the AC Sweep and Noise Analysis dialog box. Since our interest will be in
the low-frequency range, we will choose a range of 1 Hz (0 Hz is an invalid entry)
to 100 Hz. If you want a frequency range starting close to 0 Hz, you would have to
choose a frequency such as 0.001 Hz or something small enough not to be noticeable
on the plot. The Total Pts.: will be set at 1000 for a good continuous plot, the Start
Freq.: at 1 Hz, and the End Freq.: at 100 Hz. The AC Sweep Type will be left on
Linear. A simulation followed by Trace-Add-V(RL:1) will result in the desired plot.
However, the computer has selected a log scale for the horizontal axis that extends
from 1 Hz to 1 kHz even though we requested a linear scale. If we choose Plot-XAxis Settings-Linear-OK, we will get a linear plot to 120 Hz, but the curve of interest is all in the low end—the log axis obviously provided a better plot for our region of interest. Returning to Plot-X-Axis Settings and choosing Log, we return to
the original plot. Our interest only lies in the region of 1 to 100 Hz, so the remaining frequencies to 1 kHz should be removed with Plot-X-Axis Settings-User Defined-1Hz to 100Hz-OK. The vertical axis also goes to 60 mV, and we want to limit
to 50 mV for this frequency range. This is accomplished with Plot-Y-Axis SettingsUser Defined-0V to 50mV-OK, after which the pattern of Fig. 11.26 will be obtained.
512
Chapter 11
BJT and JFET Frequency Response
f
Figure 11.26 Low-frequency
response due to CS.
Note how closely the curve approaches 50 mV in this range. The cutoff level is
determined by 0.707(49.67 mV) 35.12 mV, which can be found by clicking the
Toggle cursor icon and moving the intersection up the graph until the 35.177-mV
level is reached for A1. At this point, the frequency of the horizontal axis can be read
as 6.74 Hz, comparing very well to the predicted value of 6.86 Hz. Note that A2 remains at the lowest level of the plot, at 1 Hz.
CC
1
fLC 2 (RC RL)CC
1
(6.28)(4 k 2.2 k)(1 F)
25.68 Hz
To investigate the effects of CC on the lower cutoff frequency, both CS and CE must
be set to 1 Farad as described above. Following the procedure outlined above will result in the plot of Fig. 11.27, with a cutoff frequency of 25.58 Hz, providing a close
match with the calculated level of 25.68 Hz.
Figure 11.27 Low-frequency
response due to CC.
513
f
CE
Rs RsR1R2 1 k40 k10 k 0.889 k
Re RE
Rs
15.76 r 2 k10
0
0.889 k
e
2 k(8.89 15.76 ) 2 k24.65 24.35 1
106
1
fLE 327 Hz
2 ReCE
3058.36
(6.28)(24.35 )(20 F)
The effect of CE can be examined using PSpice Windows by setting both CS and
CC to 1 Farad. In addition, since the frequency range is greater, the start frequency
has to be changed to 10 Hz and the final frequency to 1 kHz. The result is the plot
of Fig. 11.28, with a cutoff frequency of 321.17 Hz, providing a close match with the
calculated value of 327 Hz.
Figure 11.28 Low-frequency response due to CE.
The fact that fLE is significantly higher than fLS or fLC suggests that it will be the
predominant factor in determining the low-frequency response for the complete system. To test the accuracy of our hypothesis, the network is simulated with all the initial values of capacitance level to obtain the results of Fig. 11.29. Note the strong similarity with the waveform of Fig. 11.28, with the only visible difference being the
higher gain at lower frequencies on Fig. 11.28. Without question, the plot supports
the fact that the highest of the low cutoff frequencies will have the most impact on
the low cutoff frequency for the system.
(b) It was mentioned earlier that dB plots are usually normalized by dividing the voltage gain Av by the magnitude of the midband gain. For Fig. 11.16, the magnitude
of the midband gain is 51.21, and naturally the ratio Av /Avmid will be 1 in the
midband region. The result is a 0-dB asymptote in the midband region as shown
in Fig. 11.30. Defining fLE as our lower cutoff frequency f1, an asymptote at 6
dB/octave can be drawn as shown in Fig. 11.30 to form the Bode plot and our
514
Chapter 11
BJT and JFET Frequency Response
f
Figure 11.29 Low-frequency response due to CS, CE, and CC.
envelope for the actual response. At f1, the actual curve is 3 dB down from the
midband level as defined by the 0.707AVmid level, permitting a sketch of the actual frequency response curve as shown in Fig. 11.30. A 6-dB/octave asymptote was drawn at each frequency defined in the analysis above to demonstrate
clearly that it is fLE for this network that will determine the 3-dB point. It is not
until about 24 dB that fLC begins to affect the shape of the envelope. The magnitude plot shows that the slope of the resultant asymptote is the sum of the asymptotes having the same sloping direction in the same frequency interval. Note
in Fig. 11.30 that the slope has dropped to 12 dB/octave for frequencies less
Figure 11.30 Low-frequency plot for the network of
Example 11.9.
11.6 Low-Frequency Response—BJT Amplifier
515
f
fL
C
-6dB/octave
-20dB/decade
-12dB/octave
-20dB/decade
Figure 11.31 dB plot of the low-frequency response of the BJT amplifier of Fig. 11.25.
than fLC and could drop to 18 dB/octave if the three defined cutoff frequencies
of Fig. 11.30 were closer together.
Using PROBE, a plot of 20 log10Av /Avmid Av /AvmiddB can be obtained by
recalling that if Vs 1 mV, the magnitude of Av /Avmid is the same as Vo /Avmid
since Vo will have the same numerical value as Av. The required Trace Expression, which is entered on the bottom of the Add Traces dialog box, appears on
the horizontal axis of Fig. 11.31. The plot clearly reveals the change in slope of
the asymptote at fLC and how the actual curve follows the envelope created by the
Bode plot. In addition, note the 3-dB drop at f1.
Keep in mind as we proceed to the next section that the analysis of this section
is not limited to the network of Fig. 11.16. For any transistor configuration it is simply necessary to isolate each R-C combination formed by a capacitive element and
determine the break frequencies. The resulting frequencies will then determine
whether there is a strong interaction between capacitive elements in determining the
overall response and which element will have the greatest impact on establishing the
lower cutoff frequency. In fact, the analysis of the next section will parallel this section as we determine the low cutoff frequencies for the FET amplifier.
11.7 LOW-FREQUENCY RESPONSE — FET
AMPLIFIER
The analysis of the FET amplifier in the low-frequency region will be quite similar
to that of the BJT amplifier of Section 11.6. There are again three capacitors of primary concern as appearing in the network of Fig. 11.32: CG, CC, and CS. Although
Fig. 11.32 will be used to establish the fundamental equations, the procedure and conclusions can be applied to most FET configurations.
516
Chapter 11
BJT and JFET Frequency Response
f
Figure 11.32 Capacitive elements that affect the low-frequency response of a JFET amplifier.
CG
For the coupling capacitor between the source and the active device, the ac equivalent network will appear as shown in Fig. 11.33. The cutoff frequency determined by
CG will then be
1
fLG 2 (Rsig Ri)CG
(11.34)
which is an exact match of Eq. (11.26). For the network of Fig. 11.32,
Ri RG
+
Vs
–
Rsig
(11.35)
CG
System
Ri
Figure 11.33 Determining the
effect of CG on the low-frequency
response.
Typically, RG Rsig, and the lower cutoff frequency will be determined primarily by
RG and CG. The fact that RG is so large permits a relatively low level of CG while
maintaining a low cutoff frequency level for fLG.
CC
For the coupling capacitor between the active device and the load the network of Fig.
11.34 will result, which is also an exact match of Fig. 11.19. The resulting cutoff
frequency is
1
fLC 2 (Ro RL)CC
(11.36)
Ro RDrd
(11.37)
For the network of Fig. 11.32,
11.7 Low-Frequency Response—FET Amplifier
517
f
CC
RL
System
Ro
System
Reg
Figure 11.34 Determining the effect of CC on
the low-frequency response.
CS
Figure 11.35 Determining the effect of CS
on the low-frequency response.
CS
For the source capacitor CS, the resistance level of importance is defined by Fig. 11.35.
The cutoff frequency will be defined by
1
fLS 2 ReqCS
(11.38)
For Fig. 11.32, the resulting value of Req:
RS
Req 1 RS(1 gmrd)/(rd RDRL)
(11.39)
which for rd becomes
1
Req RS
gm
EXAMPLE 11.10
(11.40)
(a) Determine the lower cutoff frequency for the network of Fig. 11.32 using the following parameters:
CG 0.01 F,
CC 0.5 F,
CS 2 F
Rsig 10 k,
RG 1 M,
RD 4.7 k,
IDSS 8mA,
VP 4 V
rd ,
RS 1 k,
RL 2.2 k
VDD 20 V
(b) Sketch the frequency response using a Bode plot.
Solution
(a) DC Analysis: Plotting the transfer curve of ID IDSS(1 VGS/VP)2 and superimposing the curve defined by VGS IDRS will result in an intersection at VGSQ 2
V and IDQ 2 mA. In addition,
2IDSS
2(8 mA)
gm0 4 mS
VP
4V
VGSQ
2 V
gm gm0 1 4 mS 1 2 mS
VP
4 V
CG
Eq. (11.34):
518
Chapter 11
1
fLG 15.8 Hz
2 (10 k 1 M)(0.01 F)
BJT and JFET Frequency Response
f
CC
Eq. (11.36):
1
fLC 46.13 Hz
2 (4.7 k 2.2 k)(0.5 F)
CS
1
1
Req RS 1 k 1 k0.5 k 333.33 gm
2 mS
Eq. (11.38):
1
fLS 238.73 Hz
2 (333.33 )(2 F)
Since fLS is the largest of the three cutoff frequencies, it defines the low cutoff frequency for the network of Fig. 11.32.
(b) The midband gain of the system is determined by
Vo
Avmid gm(RDRL) (2 mS)(4.7 k2.2 k)
Vi
(2 mS)(1.499 k)
3
Using the midband gain to normalize the response for the network of Fig. 11.32 will
result in the frequency plot of Fig. 11.36.
Figure 11.36 Low-frequency
response for the JFET configuration of Example 11.10.
Using PSpice Windows, the network will appear as shown in Fig. 11.37, with the
JFET parameters Beta set at 0.5mA/V2 and Vto at 4 V (all others set to zero) and
the frequency of interest at a midband value of 10 kHz. The resulting dc levels confirm that VGS is 2 V and place VD at 10.60 V, which should be right in the middle
of the linear active region since VGS 1/2(VD 4 V) and VDS 1/2(VDD 20 V).
The 0-V levels clearly reveal that the capacitors have isolated the transistor for the dc
biasing. The ac response results in an ac level of 2.993 mV across the load for a gain
of 2.993, which is essentially equal to the calculated gain of 3.
11.7 Low-Frequency Response—FET Amplifier
519
f
Figure 11.37 Schematic network for Example 11.10.
Returning to Analysis and choosing Automatically run Probe after simulation
followed by Setup-AC Sweep-Decade-Pts/Decade 1000, Start Freq.: 10Hz, and
End Freq.: 10 kHz will setup Simulation-Trace-Add-Trace Expression: DB
(V(RL:1)/2.993mV)-OK, which will result in the plot of Fig. 11.38, with a low cutoff frequency of 227.5 Hz primarily determined by the source capacitance.
Figure 11.38 dB response for
the low-frequency region in the
network of Example 11.10.
11.8 MILLER EFFECT CAPACITANCE
In the high-frequency region, the capacitive elements of importance are the interelectrode (between terminals) capacitances internal to the active device and the wiring
capacitance between leads of the network. The large capacitors of the network that
controlled the low-frequency response have all been replaced by their short-circuit
equivalent due to their very low reactance levels.
520
Chapter 11
BJT and JFET Frequency Response
f
For inverting amplifiers (phase shift of 180° between input and output resulting
in a negative value for Av), the input and output capacitance is increased by a capacitance level sensitive to the interelectrode capacitance between the input and output
terminals of the device and the gain of the amplifier. In Fig. 11.39, this “feedback”
capacitance is defined by Cf.
Figure 11.39 Network employed in
the derivation of an equation for the
Miller input capacitance.
Applying Kirchhoff’s current law gives
Ii I1 I2
Using Ohm’s law yields
Vi
Ii ,
Zi
Vi
I1 Ri
Vi Vo
Vi AvVi
(1 Av)Vi
I2 XCf
XCf
XCf
and
Substituting, we obtain
(1 Av)Vi
Vi
Vi
XCf
Zi
Ri
1
1
1
Zi
Ri
XCf /(1 Av)
but
XCf
1
XCM
(1 Av)Cf
1 Av







and
CM
1
1
1
Zi
Ri
XCM
and
establishing the equivalent network of Fig. 11.40. The result is an equivalent input
impedance to the amplifier of Fig. 11.39 that includes the same Ri that we have dealt
with in previous chapters, with the addition of a feedback capacitor magnified by the
Ii
+
Vi
–
Zi
Ri
CM = (1 − Aυ ) Cf
Figure 11.40 Demonstrating
the impact of the Miller effect
capacitance.
11.8 Miller Effect Capacitance
521
f
gain of the amplifier. Any interelectrode capacitance at the input terminals to the amplifier will simply be added in parallel with the elements of Fig. 11.40.
In general, therefore, the Miller effect input capacitance is defined by
CMi (1 Av)Cf
(11.41)
This shows us that:
For any inverting amplifier, the input capacitance will be increased by a
Miller effect capacitance sensitive to the gain of the amplifier and the interelectrode capacitance connected between the input and output terminals of the
active device.
The dilemma of an equation such as Eq. (11.41) is that at high frequencies the
gain Av will be a function of the level of CMi. However, since the maximum gain is
the midband value, using the midband value will result in the highest level of CMi and
the worst-case scenario. In general, therefore, the midband value is typically employed
for Av in Eq. (11.41).
The reason for the constraint that the amplifier be of the inverting variety is now
more apparent when one examines Eq. (11.41). A positive value for Av would result
in a negative capacitance (for Av 1).
The Miller effect will also increase the level of output capacitance, which must
also be considered when the high-frequency cutoff is determined. In Fig. 11.41, the
parameters of importance to determine the output Miller effect are in place. Applying Kirchhoff’s current law will result in
Io I1 I2
with
Vo
I1 Ro
and
Vo Vi
I2 XCf
The resistance Ro is usually sufficiently large to permit ignoring the first term of the
equation compared to the second term and assuming that
Vo Vi
Io XCf
Substituting Vi Vo /Av from Av Vo /Vi will result in
Vo Vo/Av
Vo(1 1/Av)
Io XCf
XCf
and
1 1/Av
Io
XCf
Vo
Figure 11.41 Network employed in
the derivation of an equation for the
Miller output capacitance.
522
Chapter 11
BJT and JFET Frequency Response
or
f
XCf
Vo
1
1
Io
1 1/Av
Cf (1 1/Av)
CMo
resulting in the following equation for the Miller output capacitance:
1
CMo 1 Cf
Av
For the usual situation where Av
(11.42a)
1, Eq. (11.42a) reduces to
CMo Cf
Av
(11.42b)
1
Examples in the use of Eq. (11.42) will appear in the next two sections as we investigate the high-frequency responses of BJT and FET amplifiers.
11.9 HIGH-FREQUENCY RESPONSE — BJT
AMPLIFIER
At the high-frequency end, there are two factors that will define the 3-dB point: the
network capacitance (parasitic and introduced) and the frequency dependence of
hfe().
Network Parameters
In the high-frequency region, the RC network of concern has the configuration appearing in Fig. 11.42. At increasing frequencies, the reactance XC will decrease in
magnitude, resulting in a shorting effect across the output and a decrease in gain. The
derivation leading to the corner frequency for this RC configuration follows along
similar lines to that encountered for the low-frequency region. The most significant
difference is in the general form of Av appearing below:
1
Av 1 j( f/f2)
(11.43)
which results in a magnitude plot such as shown in Fig. 11.43 that drops off at
6 dB/octave with increasing frequency. Note that f2 is in the denominator of the
frequency ratio rather than the numerator as occurred for f1 in Eq. (11.21).
In Fig. 11.44, the various parasitic capacitances (Cbe, Cbc, Cce) of the transistor
Figure 11.42 R-C combination
that will define a high cutoff frequency.
Figure 11.43 Asymptotic plot
as defined by Eq. (11.43).
have been included with the wiring capacitances (CWi, CWo) introduced during construction. The high-frequency equivalent model for the network of Fig. 11.44 appears
in Fig. 11.45. Note the absence of the capacitors Cs, CC, and CE, which are all assumed to be in the short-circuit state at these frequencies. The capacitance Ci includes
11.9 High-Frequency Response—BJT Amplifier
523
f
VCC
RC
R1
Cbc
CC
Cs
Cce
+
Rs
CWo
Vi
+
RL
Cbe
CWi
CE
RE
Vs
–
–
Figure 11.44 Network of Fig.
11.16 with the capacitors that
affect the high-frequency
response.
R2
Co = CWo + Cce + CMo
Ci = CWi + Cbe + CMi
Ib
Rs
Vo
+
Th1
R1 R2
Vs
Ri
Th2
Ci
β Ib
ro
RC
RL
Co
–
Figure 11.45 High-frequency
ac equivalent model for the
network of Fig. 11.44.
the input wiring capacitance CWi, the transition capacitance Cbe, and the Miller capacitance CMi. The capacitance Co includes the output wiring capacitance CWo, the
parasitic capacitance Cce, and the output Miller capacitance CMo. In general, the capacitance Cbe is the largest of the parasitic capacitances, with Cce the smallest. In fact,
most specification sheets simply provide the levels of Cbe and Cbc and do not include
Cce unless it will affect the response of a particular type of transistor in a specific area
of application.
Determining the Thévenin equivalent circuit for the input and output networks of
Fig. 11.45 will result in the configurations of Fig. 11.46. For the input network, the
3-dB frequency is defined by
1
fHi 2 RTh1Ci
RTh = Rs R1 R2  Ri
RTh = RC RL  ro
1
2
+
+
Ci
1
–
(a)
524
Chapter 11
ETh
Co
2
–
ETh
(11.44)
(b)
BJT and JFET Frequency Response
Figure 11.46 Thévenin circuits
for the input and output networks
of the network of Fig. 11.45.
f
with
RTh1 RsR1R2Ri
(11.45)
and
Ci CWi Cbe CMi CWi Cbe (1 Av)Cbc
(11.46)
At very high frequencies, the effect of Ci is to reduce the total impedance of the parallel combination of R1, R2, Ri, and Ci in Fig. 11.45. The result is a reduced level of
voltage across Ci, a reduction in Ib, and a gain for the system.
For the output network,
1
fHo 2 RTh2Co
(11.47)
with
RTh2 RCRLro
(11.48)
and
Co CWo Cce CMo
(11.49)
At very high frequencies, the capacitive reactance of Co will decrease and consequently reduce the total impedance of the output parallel branches of Fig. 11.45. The
net result is that Vo will also decline toward zero as the reactance XC becomes smaller.
The frequencies fHi and fHo will each define a 6-dB/octave asymptote such as depicted in Fig. 11.43. If the parasitic capacitors were the only elements to determine
the high cutoff frequency, the lowest frequency would be the determining factor. However, the decrease in hfe (or ) with frequency must also be considered as to whether
its break frequency is lower than fHi or fHo.
hfe (or ) Variation
The variation of hfe (or ) with frequency will approach, with some degree of accuracy, the following relationship:
hfemid
hfe 1 j(f/f)
(11.50)
The use of hfe rather than in some of this descriptive material is due primarily
to the fact that manufacturers typically use the hybrid parameters when covering this
issue in their specification sheets, and so on.
The only undefined quantity, f, is determined by a set of parameters employed
in the hybrid or Giacoletto model frequently applied to best represent the transistor in the high-frequency region. It appears in Fig. 11.47. The various parameters warrant a moment of explanation. The resistance rbb includes the base contact, base bulk,
and base spreading resistance. The first is due to the actual connection to the base.
The second includes the resistance from the external terminal to the active region of
rb'c
b'
B
Ib
rbb'
1
rb'e = g
b'e
E
C
Cb'c
I'b
Cb'e
rce =
1
hoe
gm Vb'e = gm rb'e I'b ≅ hfe
I'
mid b
E
11.9 High-Frequency Response—BJT Amplifier
Figure 11.47 Giacoletto (or
hybrid ) high-frequency transistor
small-signal ac equivalent circuit.
525
f
the transistors, while the last is the actual resistance within the active base region. The
resistances rbe, rce, and rbc are the resistances between the indicated terminals when
the device is in the active region. The same is true for the capacitances Cbc and Cbe,
although the former is a transition capacitance while the latter is a diffusion capacitance. A more detailed explanation of the frequency dependence of each can be found
in a number of readily available texts.
In terms of these parameters,
gbe
f (sometimes appearing as fhfe ) 2 (Cbe Cbc)
(11.51)
or since the hybrid parameter hfe is related to gbe through gm hfemid gbe,
gm
1
f hfemid 2 (Cbe Cbc)
(11.52)
Taking it a step further,
hfemid
mid
1
1
gm hfemid gbe hfemid hie
rbe
midre
re
and using the approximations
Cbe Cbe
Cbc Cbc
and
will result in the following form for Eq. (11.50):
1
f 2 midre(Cbe Cbc)
(11.53)
Equation (11.53) clearly reveals that since re is a function of the network design:
f is a function of the bias conditions.
The basic format of Eq. (11.50) is exactly the same as Eq. (11.43) if we extract
the multiplying factor hfemid, revealing that hfe will drop off from its midband value
with a 6-dB/octave slope as shown in Fig. 11.48. The same figure has a plot of h f b
(or ) versus frequency. Note the small change in h f b for the chosen frequency range,
revealing that the common-base configuration displays improved high-frequency characteristics over the common-emitter configuration. Recall also the absence of the
Miller effect capacitance due to the noninverting characteristics of the common-base
configuration. For this very reason, common-base high-frequency parameters rather
than common-emitter parameters are often specified for a transistor—especially those
designed specifically to operate in the high-frequency regions.
The following equation permits a direct conversion for determining f if f and are specified.
f f(1 )
(11.54)
A quantity called the gain–bandwidth product is defined for the transistor by the
condition
hfemid
1
1 j( f/f)
526
Chapter 11
BJT and JFET Frequency Response
f
 hfe , hfb
 hfe
hfe
mid
40 dB
Midband value for hfe
1.0
0.707
−3 dB
30 dB
 hfe
20 dB
−6
/o
dB
cta
pe
0 dB
−3 dB
slo
 hfe = 1
ve
10 dB
Midband value for hfb
 hfb
−10 dB
fβ , ( fh )
fe
−20 dB
0.1 MHz
1.0 MHz
5fβ
100.0 MHz
10.0 MHz
fh , ( f α )
fT
fb
1 kMHz
10 kMHz
Figure 11.48 hfe and hfb versus frequency in the high-frequency region.
so that
hfemid
hfedB 20 log10 20 log10 1 0 dB
1 j( f/f)
The frequency at which hfedB 0 dB is clearly indicated by fT in Fig. 11.48. The
magnitude of hfe at the defined condition point ( fT f) is given by
hfemid
hfemid
1
fT/f
1
2
( f
T/f)



( BW)
so that
or
with
fT hfemid f
(gain–bandwidth product)
(11.55)
fT mid f
(11.56)
fT
f mid
(11.57)
Substituting Eq. (11.53) for f in Eq. (11.55) gives
1
fT mid 2 midre(Cbe Cbc)
and
1
fT 2 re(Cbe Cbc)
(11.58)
11.9 High-Frequency Response—BJT Amplifier
527
f
EXAMPLE 11.11
For the network of Fig. 11.44 with the same parameters as in Example 11.9, that is,
Rs 1 k, R1 40 k, R2 10 k, RE 2 k, RC 4 k, RL 2.2 k
Cs 10 F, CC 1 F, CE 20 F
100, ro , VCC 20 V
with the addition of
Cbe 36 pF, Cbc 4 pF, Cce 1 pF, CWi 6 pF, CWo 8 pF
(a) Determine fHi and fHo.
(b) Find f and fT.
(c) Sketch the frequency response for the low- and high-frequency regions using the
results of Example 11.9 and the results of parts (a) and (b).
(d) Obtain a PROBE response for the full frequency spectrum and compare with the
results of part (c).
Solution
(a) From Example 11.9:
Ri 1.32 k,
Avmid(amplifier) 90
RTh1 RsR1R2Ri 1 k40 k10 k1.32 k
and
0.531 k
Ci CWi Cbe (1 Av)Cbe
with
6 pF 36 pF [1 (90)]4 pF
406 pF
1
1
fHi 2 RTh1Ci
2 (0.531 k)(406 pF)
738.24 kHz
RTh2 RCRL 4 k2.2 k 1.419 k
1
Co CWo Cce CMo 8 pF 1 pF 1 4 pF
90
13.04 pF
1
1
fHo 2 RTh2Co
2 (1.419 k)(13.04 pF)
8.6 MHz
(b) Applying Eq. (11.53) gives
1
f 2 midre(Cbe Cbc)
1
1
2 (100)(15.76 )(36 pF 4 pF)
2 (100)(15.76 )(40 pF)
2.52 MHz
fT mid f (100)(2.52 MHz)
252 MHz
528
Chapter 11
BJT and JFET Frequency Response
f
(c) See Fig. 11.49. Both f and fHo will lower the upper cutoff frequency below the
level determined by fHi. f is closer to fHi and therefore will have a greater impact
than fHo. In any event, the bandwidth will be less than that defined solely by fHi.
In fact, for the parameters of this network the upper cutoff frequency will be relatively close to 600 kHz.
Aυ
Aυ
1
mid dB
fL 10
s
fL
C
100
fL
E
1 kHz
10 kHz
100 kHz
fHi
1 MHz
0
–3
fβ
10 MHz
100 MHz
f (log scale)
fHo
f2
f1
–5
BW
−6 dB/octave
–10
–15
+20 dB/decade
–20
−12 dB/octave
Figure 11.49 Full frequency
response for the network of
Fig. 11.44.
–25
In general, therefore, the lowest of the upper-cutoff frequencies defines a maximum possible bandwidth for a system.
(d) In order to obtain a PSpice analysis for the full frequency range, the parasitic capacitances have to be added to the network as shown in Fig. 11.50.
Figure 11.50 Network of Figure 11.25 with parasitic capacitances in place.
11.9 High-Frequency Response—BJT Amplifier
529
f
An Analysis will result in the plot of Fig. 11.51 using the Trace Expression appearing at the bottom of the plot. The vertical scale was changed from 60 to 0 dB
to 30 to 0 dB to highlight the area of interest using the Y-Axis Settings. The low
cutoff frequency of 324 Hz is as determined primarily by fLE, and the high cutoff frequency is near 667kHz. Even though fHo is more than a decade higher than fHi, it will
have an impact on the high cutoff frequency. In total, however, the PSpice analysis
has been a welcome verification of the hand-written approach.
Figure 11.51 Full frequency response for the network of Fig. 11.50.
11.10 HIGH-FREQUENCY RESPONSE —
FET AMPLIFIER
The analysis of the high-frequency response of the FET amplifier will proceed in a
very similar manner to that encountered for the BJT amplifier. As shown in Fig. 11.52,
there are interelectrode and wiring capacitances that will determine the high-frequency
characteristics of the amplifier. The capacitors Cgs and Cgd typically vary from 1
to 10 pF, while the capacitance Cds is usually quite a bit smaller, ranging from 0.1 to
1 pF.
Since the network of Fig. 11.52 is an inverting amplifier, a Miller effect capacitance will appear in the high-frequency ac equivalent network appearing in Fig. 11.53.
At high frequencies, Ci will approach a short-circuit equivalent and Vgs will drop in
value and reduce the overall gain. At frequencies where Co approaches its shortcircuit equivalent, the parallel output voltage Vo will drop in magnitude.
The cutoff frequencies defined by the input and output circuits can be obtained
by first finding the Thévenin equivalent circuits for each section as shown in Fig.
11.54. For the input circuit,
530
Chapter 11
BJT and JFET Frequency Response
f
VDD
RD
CC
Vo
Cgd
Cds
+
Rsig
CG
Vs
CW o
CW i
RG
RL
Cgs
–
Figure 11.52 Capacitive elements that affect the high frequency response of a JFET amplifier.
CS
RS
Rsig
Vo
+
+
Vs
Ci
RG
Vgs
gm Vgs
rd
RD
RL
Co
–
Th1
–
Figure 11.53 High-frequency
ac equivalent circuit for
Fig. 11.52.
Th2
1
fHi 2 RTh1Ci
(11.59)
and
RTh1 RsigRG
(11.60)
with
Ci CWi Cgs CMi
(11.61)
and
CMi (1 Av)Cgd
(11.62)
1
fHo 2 RTh2Co
(11.63)
and for the output circuit,
Figure 11.54 The Thévenin
equivalent circuits for the
(a) input circuit and (b) output
circuit.
11.10 High-Frequency Response—FET Amplifier
531
f
EXAMPLE 11.12
with
RTh2 RDRLrd
and
Co CWo Cds CMo
and
1
CMo 1 Cgd
Av
(11.64)
(11.65)
(a) Determine the high cutoff frequencies for the network of Fig. 11.52 using the
same parameters as Example 11.10:
CG 0.01 F,
Rsig 10 k,
RG 1 M,
IDSS 8 mA,
CC 0.5 F,
RD 4.7 k,
VP 4 V,
CS 2 F
RS 1 k,
rd ,
RL 2.2 k
VDD 20 V
with the addition of
Cgd 2 pF,
Cgs 4 pF,
Cds 0.5 pF,
CWi 5 pF,
CWo 6 pF
(b) Review a PROBE response for the full frequency range and note whether it supports the conclusions of Example 11.10 and the calculations above.
Solution
(a) RTh1 RsigRG 10 k1 M 9.9 k
From Example 11.10, Av 3.
Ci CWi Cgs (1 Av)Cgd
5 pF 4 pF (1 3)2 pF
9 pF 8 pF
17 pF
1
fHi 2 RTh1Ci
1
945.67 kHz
2 (9.9 k)(17 pF)
RTh2 RD RL
4.7 k 2.2 k
1.5 k
1
Co CWo Cds CMo 6 pF 0.5 pF 1 2 pF 9.17 pF
3
1
fHo 11.57 MHz
2 (1.5 k)(9.17 pF)
The results above clearly indicate that the input capacitance with its Miller effect capacitance will determine the upper cutoff frequency. This is typically the case due to
the smaller value of Cds and the resistance levels encountered in the output circuit.
(b) Using PSpice Windows, the schematic for the network will appear as shown in
Fig. 11.55.
532
Chapter 11
BJT and JFET Frequency Response
f
Figure 11.55 Network of Figure 11.52 with assigned values.
Under Analysis, the AC Sweep is set to Decade with Pts/Decade at 1000, Start
Freq.: at 10 Hz, and End Freq.: at 10 MHz. Under the Add Traces dialog box, the
Trace Expression is entered as DB(V(RL:1)/2.993mV), and the plot of Fig. 11.56
is obtained. Just for a moment, consider how much time it must have taken to obtain
a plot such as in Fig. 11.56 without computer methods for a network as complicated
as Fig. 11.55. Often, we forget how computer systems have helped us through some
painstaking, lengthy, and boring series of calculations.
Figure 11.56 Frequency response for the network of Example 11.12.
Using the cursor, we find the lower and upper cutoff frequencies to be 225 Hz
and 921 kHz, respectively, providing a nice match with the calculated values.
11.10 High-Frequency Response—FET amplifier
533
f
Even though the analysis of the past few sections has been limited to two configurations, the exposure to the general procedure for determining the cutoff frequencies should support the analysis of any other transistor configuration. Keep in
mind that the Miller capacitance is limited to inverting amplifiers and that f is significantly greater than f if the common-base configuration is encountered. There is
a great deal more literature on the analysis of single-stage amplifiers that goes beyond the coverage of this chapter. However, the content of this chapter should provide a firm foundation for any future analysis of frequency effects.
11.11 MULTISTAGE FREQUENCY EFFECTS
For a second transistor stage connected directly to the output of a first stage, there
will be a significant change in the overall frequency response. In the high-frequency
region, the output capacitance Co must now include the wiring capacitance (CW1), parasitic capacitance (Cbe), and Miller capacitance (CMi) of the following stage. Further,
there will be additional low-frequency cutoff levels due to the second stage that will
further reduce the overall gain of the system in this region. For each additional stage,
the upper cutoff frequency will be determined primarily by that stage having the lowest cutoff frequency. The low-frequency cutoff is primarily determined by that stage
having the highest low-frequency cutoff frequency. Obviously, therefore, one poorly
designed stage can offset an otherwise well-designed cascaded system.
The effect of increasing the number of identical stages can be clearly demonstrated by considering the situations indicated in Fig. 11.57. In each case, the upper
and lower cutoff frequencies of each of the cascaded stages are identical. For a single stage, the cutoff frequencies are f1 and f2 as indicated. For two identical stages in
cascade, the drop-off rate in the high- and low-frequency regions has increased to
12 dB/octave or 40 dB/decade. At f1 and f2, therefore, the decibel drop is now
6 dB rather than the defined band frequency gain level of 3 dB. The 3-dB
point has shifted to f1 and f2 as indicated, with a resulting drop in the bandwidth.
A 18-dB/octave or 60-dB/decade slope will result for a three-stage system of identical stages with the indicated reduction in bandwidth (f 1 and f 2).
Assuming identical stages, an equation for each band frequency as a function of
the number of stages (n) can be determined in the following manner: For the lowfrequency region,
Avlow, (overall) Av1 Av2 Av3
low
low
low
Avn
low
Figure 11.57 Effect of an increased number of stages on the cutoff frequencies
and the bandwidth.
534
Chapter 11
BJT and JFET Frequency Response
f
but since each stage is identical, Av1 Av2 etc. and
low
low
Avlow, (overall) (Av1 )n
low
or
Avlow
Av1
(overall) Avmid
Avmid
low
n
1
n
(1 jf1/f )
Setting the magnitude of this result equal to 1/2 (3 dB level) results in
1
1
2 n 1
f
( /f
)]
2
1
1
or
f1
1 f 1
2 1/2 n
f1
1 f 1
2 n 1/2
2
21/n
so that
f1
1 f 1
and
f1
1 f 1
with the result that
f1
f 1 2 n
2
(2)1/2
(11.66)
In a similar manner, it can be shown that for the high-frequency region,
1/
n
)
1f2
f2 (2
(11.67)
Note the presence of the same factor 21/
n
1 in each equation. The magnitude of
this factor for various values of n is listed below.
n
21/
n
1
2
3
4
5
0.64
0.51
0.43
0.39
For n 2, consider that the upper cutoff frequency f2 0.64f2 or 64% of the value
obtained for a single stage, while f1 (1/0.64)f1 1.56f1. For n 3, f 2 0.51f2 or
approximately 12 the value of a single stage with f1 (1/0.51)f1 1.96f1 or approximately twice the single-stage value.
For the RC-coupled transistor amplifier, if f2 f, or if they are close enough in
magnitude for both to affect the upper 3-dB frequency, the number of stages must be
increased by a factor of 2 when determining f2 due to the increased number of factors 1/(1 jf/fx).
A decrease in bandwidth is not always associated with an increase in the number
of stages if the midband gain can remain fixed and independent of the number of
stages. For instance, if a single-stage amplifier produces a gain of 100 with a bandwidth of 10,000 Hz, the resulting gain–bandwidth product is 102104 106. For a
two-stage system the same gain can be obtained by having two stages with a gain of
10 since (10 10 100). The bandwidth of each stage would then increase by a factor of 10 to 100,000 due to the lower gain requirement and fixed gain–bandwidth
product of 106. Of course, the design must be such as to permit the increased bandwidth and establish the lower gain level.
11.11 Multistage Frequency Effects
535
f
11.12 SQUARE-WAVE TESTING
A sense for the frequency response of an amplifier can be determined experimentally
by applying a square-wave signal to the amplifier and noting the output response.
The shape of the output waveform will reveal whether the high or low frequencies
are being properly amplified. The use of square-wave testing is significantly less
time-consuming than applying a series of sinusoidal signals at different frequencies
and magnitudes to test the frequency response of the amplifier.
The reason for choosing a square-wave signal for the testing process is best described by examining the Fourier series expansion of a square wave composed of a
series of sinusoidal components of different magnitudes and frequencies. The summation of the terms of the series will result in the original waveform. In other words,
even though a waveform may not be sinusoidal, it can be reproduced by a series of
sinusoidal terms of different frequencies and magnitudes.
The Fourier series expansion for the square wave of Fig. 11.58 is
4
1
1
1
v Vm sin 2 fst sin 2 (3fs)t sin 2 (5fs)t sin 2 (7fs)t
3
5
7
1
sin 2 (9fs)t 9
Figure 11.58 Square wave.
(11.68)
The first term of the series is called the fundamental term and in this case has the
same frequency, fs, as the square wave. The next term has a frequency equal to three
times the fundamental and is referred to as the third harmonic. Its magnitude is onethird the magnitude of the fundamental term. The frequencies of the succeeding terms
are odd multiples of the fundamental term, and the magnitude decreases with each
higher harmonic. Figure 11.59 demonstrates how the summation of terms of a Fourier
series can result in a nonsinusoidal waveform. The generation of the square wave of
Fig. 11.58 would require an infinite number of terms. However, the summation of just
the fundamental term and the third harmonic in Fig. 11.59a clearly results in a waveform that is beginning to take on the appearance of a square wave. Including the fifth
and seventh harmonics as in Fig. 11.59b takes us a step closer to the waveform of
Fig. 11.58.
Since the ninth harmonic has a magnitude greater than 10% of the fundamental
term [19(100%) 11.1%], the fundamental term through the ninth harmonic are the
major contributors to the Fourier series expansion of the square-wave function. It is
therefore reasonable to assume that if the application of a square wave of a particular frequency results in a nice clean square wave at the output, then the fundamental
Figure 11.59 Harmonic content
of a square wave.
536
1
sin 2 (nfs)t)
n
Chapter 11
BJT and JFET Frequency Response
f
through the ninth harmonic are being amplified without visual distortion by the amplifier. For instance, if an audio amplifier with a bandwidth of 20 kHz (audio range
is from 20 Hz to 20 kHz) is to be tested, the frequency of the applied signal should
be at least 20 kHz/9 2.22 kHz.
If the response of an amplifier to an applied square wave is an undistorted replica
of the input, the frequency response (or BW) of the amplifier is obviously sufficient
for the applied frequency. If the response is as shown in Fig. 11.60a and b, the low
frequencies are not being amplified properly and the low cutoff frequency has to be
investigated. If the waveform has the appearance of Fig. 11.60c, the high-frequency
components are not receiving sufficient amplification and the high cutoff frequency
(or BW) has to be reviewed.
v
v
T
2
0
T
3T
2
2T
t
T
2
0
(a)
3T
2
2T
t
3T
2
2T
t
(b)
v
0
T
v
T
2
T
(c)
3T
2
2T
t
0
T
2
T
(d)
Figure 11.60 (a) Poor low frequency
response; (b) very poor low-frequency
response; (c) poor high-frequency
response; (d) very poor high-frequency
response.
The actual high cutoff frequency (or BW) can be determined from the output
waveform by carefully measuring the rise time defined between 10% and 90% of the
peak value, as shown in Fig. 11.61. Substituting into the following equation will provide the upper cutoff frequency, and since BW fHi fLo fHi, the equation also
provides an indication of the BW of the amplifier.
Figure 11.61 Defining the rise
time and tilt of a square wave
response.
11.12 Square-Wave Testing
537
f
0.35
BW fHi tr
(11.69)
The low cutoff frequency can be determined from the output response by carefully measuring the tilt of Fig. 11.61 and substituting into one of the following equations:
V V
(11.70)
% tilt P% 100%
V
V V
tilt P V
(decimal form)
(11.71)
The low cutoff frequency is then determined from
P
fLo fs
EXAMPLE 11.13
(11.72)
The application of a 1-mV, 5-kHz square wave to an amplifier resulted in the output
waveform of Fig. 11.62.
(a) Write the Fourier series expansion for the square wave through the ninth harmonic.
(b) Determine the bandwidth of the amplifier.
(c) Calculate the low cutoff frequency.
Solution
4 mV
(a) vi sin 2
1
1
(5 103)t sin 2 (15 103)t sin 2 (25 103)t
3
5
1
1
sin 2 (35103)t sin 2 (45 103)t
7
9
Figure 11.62 Example 11.13
(b) tr 18 s 2 s 16 s
0.35
0.35
BW 21,875 Hz 4.4fs
16 s
tr
V V
50 mV 40 mV
(c) P 0.2
V
50 mV
P
0.2
fLo fs (5 kHz) 318.31 Hz
11.13 PSPICE WINDOWS
The computer analysis of this chapter was integrated into the chapter for emphasis
and a clear demonstration of the power of the PSpice software package. The complete frequency response of a single-stage or multistage system can be determined in
a relatively short period of time to verify theoretical calculations or provide an immediate indication of the low and high cutoff frequencies of the system. The exercises in the chapter will provide an opportunity to apply the PSpice software package to a variety of networks.
538
Chapter 11
BJT and JFET Frequency Response
f
§ 11.2 Logarithms
1.
(a) Determine the common logarithm of the following numbers: 103, 50, and 0.707.
PROBLEMS
(b) Determine the natural logarithm of the same numbers appearing in part (a).
(c) Compare the solutions of parts (a) and (b).
2.
(a) Determine the common logarithm of the number 2.2 103.
(b) Determine the natural logarithm of the number of part (a) using Eq. (11.4).
(c) Determine the natural logarithm of the number of part (a) using natural logarithms and
compare with the solution of part (b).
3.
Determine:
(a) 20 log10 480 using Eq. (11.6) and compare with 20 log10 5.
(b) 10 log10 210 using Eq. (11.7) and compare with 10 log10 0.05.
(c) log10(40)(0.125) using Eq. (11.8) and compare with log10 5.
4.
Calculate the power gain in decibels for each of the following cases.
(a) Po 100 W, Pi 5 W.
(b) Po 100 mW, Pi 5 mW.
(c) Po 100 W, Pi 20 W.
5. Determine GdBm for an output power level of 25 W.
6.
Two voltage measurements made across the same resistance are V1 25 V and V2 100 V.
Calculate the power gain in decibels of the second reading over the first reading.
7.
Input and output voltage measurements of Vi 10 mV and Vo 25 V are made. What is the
voltage gain in decibels?
*
8.
(a) The total decibel gain of a three-stage system is 120 dB. Determine the decibel gain of
each stage if the second stage has twice the decibel gain of the first and the third has 2.7
times the decibel gain of the first.
(b) Determine the voltage gain of each stage.
*
9. If the applied ac power to a system is 5 W at 100 mV and the output power is 48 W, determine:
(a) The power gain in decibels.
(b) The voltage gain in decibels if the output impedance is 40 k.
(c) The input impedance.
(d) The output voltage.
§ 11.4 General Frequency Considerations
10.
Given the characteristics of Fig. 11.63, sketch:
(a) The normalized gain.
(b) The normalized dB gain (and determine the bandwidth and cutoff frequencies).
Figure 11.63 Problem 10
Problems
539
f
§ 11.5 Low-Frequency Analysis—Bode Plot
0.068 µ F
+
Vi
–
11.
For the network of Fig. 11.64:
(a) Determine the mathematical expression for the magnitude of the ratio Vo /Vi.
(b) Using the results of part (a), determine Vo /Vi at 100 Hz, 1 kHz, 2 kHz, 5 kHz, and
10 kHz, and plot the resulting curve for the frequency range of 100 Hz to 10 kHz. Use
a log scale.
(c) Determine the break frequency.
(d) Sketch the asymptotes and locate the 3-dB point.
(e) Sketch the frequency response for Vo /Vi and compare to the results of part (b).
12.
For the network of Fig. 11.64:
(a) Determine the mathematical expression for the angle by which Vo leads Vi.
(b) Determine the phase angle at f 100 Hz, 1 kHz, 2 kHz, 5 kHz, and 10 kHz, and plot
the resulting curve for the frequency range of 100 Hz to 10 kHz.
(c) Determine the break frequency.
(d) Sketch the frequency response of for the same frequency spectrum of part (b) and compare results.
13.
(a)
(b)
(c)
(d)
+
1.2 kΩ
Vo
–
Figure 11.64 Problems 11, 12,
and 32
What frequency is 1 octave above 5 kHz?
What frequency is 1 decade below 10 kHz?
What frequency is 2 octaves below 20 kHz?
What frequency is 2 decades above 1 kHz?
§ 11.6 Low-Frequency Response — BJT Amplifier
14.
Repeat the analysis of Example 11.9 with ro 40 k. What is the effect on Avmid , fLS , fLC ,
fLE, and the resulting cutoff frequency?
15.
For
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
the network of Fig. 11.65:
Determine re.
Find Avmid Vo /Vi.
Calculate Zi.
Find AvSmid Vo /Vs.
Determine fLS, fLC, and fLE.
Determine the low cutoff frequency.
Sketch the asymptotes of the Bode plot defined by the cutoff frequencies of part (e).
Sketch the low-frequency response for the amplifier using the results of part (f).
Figure 11.65 Problems 15, 22, and 33
540
Chapter 11
BJT and JFET Frequency Response
f
*
16. Repeat Problem 15 for the emitter-stabilized network of Fig. 11.66.
Figure 11.66 Problems 16 and 23
*
17. Repeat Problem 15 for the emitter-follower network of Fig. 11.67.
14 V
120 kΩ
Vi
1 kΩ
+
Vs
Cbc = 20 pF
Cbe = 30 pF
Cce = 12 pF
β = 100
0.1 µ F
0.1 µF
Zi
CW = 8 pF
i
CWo = 10 pF
30 kΩ
2.2 kΩ
8.2 kΩ
–
Figure 11.67 Problems 17 and 24
*
18. Repeat Problem 15 for the common-base configuration of Fig. 11.68. Keep in mind that the
common-base configuration is a noninverting network when you consider the Miller effect.
Figure 11.68 Problems 18, 25, and 34
Problems
541
f
§ 11.7 Low-Frequency Response — FET Amplifier
19.
For the network of Fig. 11.69:
(a) Determine VGSQ and IDQ.
(b) Find gm0 and gm.
(c) Calculate the midband gain of Av Vo /Vi.
(d) Determine Zi.
(e) Calculate Avs Vo /Vs.
(f) Determine fLG, fLC, and fLS.
(g) Determine the low cutoff frequency.
(h) Sketch the asymptotes of the Bode plot defined by part (f).
(i) Sketch the low-frequency response for the amplifier using the results of part (f).
Figure 11.69 Problems 19, 20,
26, and 35
*
20. Repeat the analysis of Problem 19 with rd 100 k. Does it have an impact of any consequence on the results? If so, which elements?
*
21. Repeat the analysis of Problem 19 for the network of Fig. 11.70. What effect did the
voltage-divider configuration have on the input impedance and the gain Avs compared to the
biasing arrangement of Fig. 11.69?
Figure 11.70 Problems 21
and 27
§ 11.9 High-Frequency Response — BJT Amplifier
22.
542
For
(a)
(b)
(c)
the network of Fig. 11.65:
Determine fHi and fHo.
Assuming that Cbe Cbe and Cbc Cbc, find f and fT.
Sketch the frequency response for the high-frequency region using a Bode plot and determine the cutoff frequency.
Chapter 11
BJT and JFET Frequency Response
f
*
23. Repeat the analysis of Problem 22 for the network of Fig. 11.66.
*
24. Repeat the analysis of Problem 22 for the network of Fig. 11.67.
*
25. Repeat the analysis of Problem 22 for the network of Fig. 11.68.
§ 11.10 High-Frequency Response — FET Amplifier
26. For
(a)
(b)
(c)
(d)
*
the network of Fig. 11.69:
Determine gm0 and gm.
Find Av and Avs in the mid-frequency range.
Determine fHi and fHo.
Sketch the frequency response for the high-frequency region using a Bode plot and determine the cutoff frequency.
27. Repeat the analysis of Problem 26 for the network of Fig. 11.70.
§ 11.11 Multistage Frequency Effects
28. Calculate the overall voltage gain of four identical stages of an amplifier, each having a gain
of 20.
29. Calculate the overall upper 3-dB frequency for a four-stage amplifier having an individual
stage value of f2 2.5 MHz.
30. A four-stage amplifier has a lower 3-dB frequency for an individual stage of f1 40 Hz. What
is the value of f 1 for this full amplifier?
§ 11.12 Square-Wave Testing
*
31. The application of a 10-mV, 100-kHz square wave to an amplifier resulted in the output waveform of Fig. 11.71.
(a) Write the Fourier series expansion for the square wave through the ninth harmonic.
(b) Determine the bandwidth of the amplifier to the accuracy available by the waveform of
Fig. 11.71.
(c) Calculate the low cutoff frequency.
Vo (mV)
100
90
80
70
60
50
40
30
20
10
0
1
2
3
4
5
6
t ( µ s)
Figure 11.71 Problem 31
§ 11.13 PSpice Windows
32. Using PSpice Windows, determine the frequency response of Vo /Vi for the high-pass filter of
Fig. 11.64.
33. Using PSpice Windows, determine the frequency response of Vo /Vs for the BJT amplifier of
Fig. 11.65.
34. Repeat Problem 33 for the network of Fig. 11.68.
35. Repeat Problem 33 for the JFET configuration of Fig. 11.69.
*Please Note: Asterisks indicate more difficult problems.
Problems
543
CHAPTER
Linear-Digital ICs
17
17.1 INTRODUCTION
While there are many ICs containing only digital circuits and many that contain only
linear circuits, there are a number of units that contain both linear and digital circuits.
Among the linear/digital ICs are comparator circuits, digital/analog converters, interface circuits, timer circuits, voltage-controlled oscillator (VCO) circuits, and phaselocked loops (PLLs).
The comparator circuit is one to which a linear input voltage is compared to another reference voltage, the output being a digital condition representing whether the
input voltage exceeded the reference voltage.
Circuits that convert digital signals into an analog or linear voltage, and those that
convert a linear voltage into a digital value, are popular in aerospace equipment, automotive equipment, and compact disk (CD) players, among many others.
Interface circuits are used to enable connecting signals of different digital voltage
levels, from different types of output devices, or from different impedances so that
both the driver stage and the receiver stage operate properly.
Timer ICs provide linear and digital circuits to use in various timing operations,
as in a car alarm, a home timer to turn lights on or off, and a circuit in electromechanical equipment to provide proper timing to match the intended unit operation.
The 555 timer has long been a popular IC unit. A voltage-controlled oscillator provides an output clock signal whose frequency can be varied or adjusted by an input
voltage. One popular application of a VCO is in a phase-locked loop unit, as used in
various communication transmitters and receivers.
17.2 COMPARATOR UNIT OPERATION
A comparator circuit accepts input of linear voltages and provides a digital output
that indicates when one input is less than or greater than the second. A basic comparator circuit can be represented as in Fig. 17.1a. The output is a digital signal that
stays at a high voltage level when the noninverting () input is greater than the voltage at the inverting () input and switches to a lower voltage level when the noninverting input voltage goes below the inverting input voltage.
Figure 17.1b shows a typical connection with one input (the inverting input in this
example) connected to a reference voltage, the other connected to the input signal
voltage. As long as Vin is less than the reference voltage level of 2 V, the output remains at a low voltage level (near 10 V). When the input rises just above 2 V, the
721
+V
−Input
+V (+10 V)
–
Output
+Input
Vref
(+2 V)
–
Vin
+
+
−V (−10 V)
−V
Figure 17.1 Comparator unit:
(a) basic unit; (b) typical
application.
Output
(a)
(b)
output quickly switches to a high-voltage level (near 10 V). Thus the high output
indicates that the input signal is greater than 2 V.
Since the internal circuit used to build a comparator contains essentially an opamp circuit with very high voltage gain, we can examine the operation of a comparator using a 741 op-amp, as shown in Fig. 17.2. With reference input (at pin 2) set
to 0 V, a sinusoidal signal applied to the noninverting input (pin 3) will cause the output to switch between its two output states, as shown in Fig. 17.2b. The input Vi going even a fraction of a millivolt above the 0-V reference level will be amplified by
the very high voltage gain (typically over 100,000) so that the output rises to its positive output saturation level and remains there while the input stays above Vref 0 V.
When the input drops just below the 0-V reference level, the output is driven to its
lower saturation level and stays there while the input remains below Vref 0 V. Figure 17.2b clearly shows that the input signal is linear while the output is digital.
Figure 17.2 Operation of 741
op-amp as comparator.
In general use, the reference level need not be 0 V but can be any desired positive or negative voltage. Also, the reference voltage may be connected to either plus
or minus input and the input signal then applied to the other input.
Use of Op-Amp as Comparator
Figure 17.3a shows a circuit operating with a positive reference voltage connected to
the minus input and the output connected to an indicator LED. The reference voltage
level is set at
10 k
Vref (12 V) 6 V
10 k 10 k
722
Chapter 17
Linear-Digital ICs
+12 V
10 kΩ
–
741
10 kΩ
Vi
Vo
+
470 Ω
−12 V
LED
LED on when
Vi goes above
Vref (= +6 V)
LED
LED on when
Vi goes below
Vref (= +6 V)
(a)
+12 V
10 kΩ
Vi
–
741
+
10 kΩ
Vo
470 Ω
−12 V
(b)
Figure 17.3 A 741 op-amp
used as a comparator.
Since the reference voltage is connected to the inverting input, the output will switch
to its positive saturation level when the input, Vi, goes more positive than the 6-V
reference voltage level. The output, Vo, then drives the LED on as an indication that
the input is more positive than the reference level.
As an alternative connection, the reference voltage could be connected to the noninverting input as shown in Fig. 17.3b. With this connection, the input signal going
below the reference level would cause the output to drive the LED on. The LED can
thus be made to go on when the input signal goes above or below the reference level,
depending on which input is connected as signal input and which as reference input.
Using Comparator IC Units
While op-amps can be used as comparator circuits, separate IC comparator units are
more suitable. Some of the improvements built into a comparator IC are faster switching between the two output levels, built-in noise immunity to prevent the output from
oscillating when the input passes by the reference level, and outputs capable of directly driving a variety of loads. A few popular IC comparators are covered next, describing their pin connections and how they may be used.
311 COMPARATOR
The 311 voltage comparator shown in Fig. 17.4 contains a comparator circuit that
can operate as well from dual power supplies of 15 V as from a single 5-V supply (as used in digital logic circuits). The output can provide a voltage at one of two
distinct levels or can be used to drive a lamp or a relay. Notice that the output is taken
17.2 Comparator Unit Operation
723
Figure 17.4
A 311 comparator (eight-pin DIP unit).
from a bipolar transistor to allow driving a variety of loads. The unit also has balance
and strobe inputs, the strobe input allowing gating of the output. A few examples will
show how this comparator unit can be used in some common applications.
A zero-crossing detector that senses (detects) the input voltage crossing through
0 V is shown using the 311 IC in Fig. 17.5. The inverting input is connected to ground
(the reference voltage). The input signal going positive drives the output transistor on,
with the output then going low (10 V in this case). The input signal going negative
(below 0 V) will drive the output transistor off, the output then going high (to 10
V). The output is thus an indication of whether the input is above or below 0 V. When
the input is any positive voltage, the output is low, while any negative voltage will
result in the output going to a high voltage level.
Figure 17.5
Zero-crossing detector using a 311 IC.
Figure 17.6 shows how a 311 comparator can be used with strobing. In this example, the output will go high when the input goes above the reference level—but
only if the TTL strobe input is off (or 0 V). If the TTL strobe input goes high, it
drives the 311 strobe input at pin 6 low, causing the output to remain in the off state
(with output high) regardless of the input signal. In effect, the output remains high
724
Chapter 17
Linear-Digital ICs
Figure 17.6 Operation of a 311
comparator with strobe input.
unless strobed. If strobed, the output then acts normally, switching from high to low
depending on the input signal level. In operation, the comparator output will respond
to the input signal only during the time the strobe signal allows such operation.
Figure 17.7 shows the comparator output driving a relay. When the input goes below 0 V, driving the output low, the relay is activated, closing the normally open (N.O.)
contacts at that time. These contacts can then be connected to operate a large variety
of devices. For example, a buzzer or bell wired to the contacts can be driven on whenever the input voltage drops below 0 V. As long as the voltage is present at the input
terminal, the buzzer will remain off.
Figure 17.7
Operation of a 311 comparator with relay output.
339 COMPARATOR
The 339 IC is a quad comparator containing four independent voltage comparator circuits connected to external pins as shown in Fig. 17.8. Each comparator has inverting and noninverting inputs and a single output. The supply voltage applied to a
pair of pins powers all four comparators. Even if one wishes to use one comparator,
all four will be drawing power.
17.2 Comparator Unit Operation
725
Figure 17.8
IC (339).
Quad comparator
To see how these comparator circuits can be used, Fig. 17.9 shows one of the 339
comparator circuits connected as a zero-crossing detector. Whenever the input signal
goes above 0 V, the output switches to V. The input switches to V only when the
input goes below 0 V.
A reference level other than 0 V can also be used, and either input terminal could
be used as the reference, the other terminal then being connected to the input signal.
The operation of one of the comparator circuits is described next.
V + (5 V)
Vi
3
5
5.1 kΩ
+
Input, Vi
Time
2
Output, Vo
339
4
Vref = 0 V
–
12
V − (−5 V)
Vo
V+
(a)
Time
V−
726
Figure 17.9
Operation of one 339 comparator circuit as a zero-crossing detector.
Chapter 17
Linear-Digital ICs
The differential input voltage (difference voltage across input terminals) going
positive drives the output transistor off (open circuit), while a negative differential input voltage drives the output transistor on—the output then at the supply low level.
If the negative input is set at a reference level Vref, the positive input goes above
Vref and results in a positive differential input with output driven to the open-circuit
state. When the noninverting input goes below Vref, resulting in a negative differential input, the output will be driven to V.
If the positive input is set at the reference level, the inverting input going below
Vref results in the output open circuit while the inverting input going above Vref results in the output at V. This operation is summarized in Fig. 17.10.
V+
V+
Vref
+
Input
+
Input < Vref ← output open circuit
Input > Vref ← output open circuit
339
Vref
339
Input < Vref ← output =V −
–
Input
V−
← output
=V−
–
V−
(a)
Figure 17.10
input.
Input > Vref
(b)
Operation of a 339 comparator circuit with reference input: (a) minus input; (b) plus
Since the output of one of these comparator circuits is from an open-circuit collector, applications in which the outputs from more than one circuit can be wire-ORed
are possible. Figure 17.11 shows two comparator circuits connected with common
output and also with common input. Comparator 1 has a 5-V reference voltage in+9 V
4
–
5.1 kΩ
3
2
7.5 kΩ
Output
1
5
+
9.1 kΩ
Vref1 ≅ +5 V
+9 V
8.2 kΩ
Vref2 ≅ +1 V
+5 V
6
–
Output high
1
1 kΩ
2
+1 V
7
+
Output low
Output low
12
Figure 17.11 Operation of two
339 comparator circuits as a
window detector.
Input
17.2 Comparator Unit Operation
727
put connected to the noninverting input. The output will be driven low by comparator 1 when the input signal goes above 5 V. Comparator 2 has a reference voltage
of 1 V connected to the inverting input. The output of comparator 2 will be driven
low when the input signal goes below 1 V. In total, the output will go low whenever the input is below 1 V or above 5 V, as shown in Fig. 17.11, the overall operation being that of a voltage window detector. The high output indicates that the input is within a voltage window of 1 to 5 V (these values being set by the reference
voltage levels used).
17.3 DIGITAL–ANALOG CONVERTERS
Many voltages and currents in electronics vary continuously over some range of values. In digital circuitry the signals are at either one of two levels, representing the binary values of 1 or zero. An analog–digital converter (ADC) obtains a digital value
representing an input analog voltage, while a digital–analog converter (DAC) changes
a digital value back into an analog voltage.
Digital-to-Analog Conversion
LADDER NETWORK CONVERSION
Digital-to-analog conversion can be achieved using a number of different methods. One popular scheme uses a network of resistors, called a ladder network. A ladder network accepts inputs of binary values at, typically, 0 V or Vref and provides an
output voltage proportional to the binary input value. Figure 17.12a shows a ladder
network with four input voltages, representing 4 bits of digital data and a dc voltage
output. The output voltage is proportional to the digital input value as given by the
relation
D0 20 D1 21 D2 22 D3 23
Vref
Vo 24
Figure 17.12 Four-stage ladder network used as a DAC: (a) basic circuit;
(b) circuit example with 0110 input.
728
Chapter 17
Linear-Digital ICs
(17.1)
In the example shown in Fig. 17.12b, the output voltage resulting should be
01121408
Vo (16 V) 6 V
16
Therefore, 01102, digital, converts to 6 V, analog.
The function of the ladder network is to convert the 16 possible binary values
from 0000 to 1111 into one of 16 voltage levels in steps of Vref/16. Using more sections of ladder allows having more binary inputs and greater quantization for each
step. For example, a 10-stage ladder network could extend the number of voltage steps
or the voltage resolution to Vref/210 or Vref/1024. A reference voltage of Vref 10 V
would then provide output voltage steps of 10 V/1024 or approximately 10 mV. More
ladder stages provide greater voltage resolution. In general, the voltage resolution for
n ladder stages is
V ef
r
2n
(17.2)
Figure 17.13 shows a block diagram of a typical DAC using a ladder network.
The ladder network, referred in the diagram as an R-2R ladder, is sandwiched between the reference current supply and current switches connected to each binary input, the resulting output current proportional to the input binary value. The binary input turns on selected legs of the ladder, the output current being a weighted summing
of the reference current. Connecting the output current through a resistor will produce an analog voltage, if desired.
Digital inputs
Current switches
Io
R-2 R ladder
Vref
Reference
current
Figure 17.13 DAC IC using
R-2R ladder network.
Analog-to-Digital Conversion
DUAL-SLOPE CONVERSION
A popular method for converting an analog voltage into a digital value is the dualslope method. Figure 17.14a shows a block diagram of the basic dual-slope converter.
The analog voltage to be converted is applied through an electronic switch to an integrator or ramp-generator circuit (essentially a constant current charging a capacitor
to produce a linear ramp voltage). The digital output is obtained from a counter operated during both positive and negative slope intervals of the integrator.
The method of conversion proceeds as follows. For a fixed time interval (usually
the full count range of the counter), the analog voltage connected to the integrator
raises the voltage at the comparator input to some positive level. Figure 17.14b shows
that at the end of the fixed time interval the voltage from the integrator is greater for
the larger input voltage. At the end of the fixed count interval, the count is set to zero
and the electronic switch connects the integrator to a reference or fixed input voltage.
17.3
Digital-Analog Converters
729
Analog input
Digital input
Integrator
Comparator
Reference
input
Count
pulses
Stop
count
Control
logic
Digital
counter
Clear
pulse
Vref
Count over
Clock
Linear
Digital
Linear/Digital
(a)
L
ar
r
ge
in
pu
ut
Inp
o
tv
lta
ge
Fixed
discharge
rate
e
tag
vol
e
put voltag
Smaller in
Smaller
digital
count
Digital
count
Larger
digital
count
Fixed time
interval
Count interval
(b)
Figure 17.14 Analog-to-digital conversion using dual-slope method: (a) logic diagram;
(b) waveform.
The integrator output (or capacitor input) then decreases at a fixed rate. The counter
advances during this time, while the integrator’s output decreases at a fixed rate until it drops below the comparator reference voltage, at which time the control logic
receives a signal (the comparator output) to stop the count. The digital value stored
in the counter is then the digital output of the converter.
Using the same clock and integrator to perform the conversion during positive and
negative slope intervals tends to compensate for clock frequency drift and integrator
accuracy limitations. Setting the reference input value and clock rate can scale the
counter output as desired. The counter can be a binary, BCD, or other form of digital counter, if desired.
LADDER-NETWORK CONVERSION
Another popular method of analog-to-digital conversion uses a ladder network
along with counter and comparator circuits (see Fig. 17.15). A digital counter advances from a zero count while a ladder network driven by the counter outputs a staircase voltage, as shown in Fig. 17.15b, which increases one voltage increment for each
count step. A comparator circuit, receiving both staircase voltage and analog input
730
Chapter 17
Linear-Digital ICs
Figure 17.15 Analog-to-digital conversion using ladder network: (a) logic diagram; (b) waveform.
voltage, provides a signal to stop the count when the staircase voltage rises above the
input voltage. The counter value at that time is the digital output.
The amount of voltage change stepped by the staircase signal depends on the number of count bits used. A 12-stage counter operating a 12-stage ladder network using
a reference voltage of 10 V would step each count by a voltage of
V f
10 V
1re
2 2.4 mV
2
4096
This would result in a conversion resolution of 2.4 mV. The clock rate of the counter
would affect the time required to carry out a conversion. A clock rate of 1 MHz operating a 12-stage counter would need a maximum conversion time of
4096 1 s 4096 s 4.1 ms
The minimum number of conversions that could be carried out each second would
then be
number of conversions 1/4.1 ms 244 conversions/second
Since on the average, with some conversions requiring little count time and others
near maximum count time, a conversion time of 4.1 ms/2 2.05 ms would be needed,
and the average number of conversions would be 2 244 488 conversions/second.
A slower clock rate would result in fewer conversions per second. A converter using
fewer count stages (and less conversion resolution) would carry out more conversions
per second. The conversion accuracy depends on the accuracy of the comparator.
17.3
Digital-Analog Converters
731
17.4 TIMER IC UNIT OPERATION
Another popular analog–digital integrated circuit is the versatile 555 timer. The IC is
made of a combination of linear comparators and digital flip-flops as described in
Fig. 17.16. The entire circuit is usually housed in an 8-pin package as specified in
Fig. 17.16. A series connection of three resistors sets the reference voltage levels to
the two comparators at 2VCC/3 and VCC/3, the output of these comparators setting or
resetting the flip-flop unit. The output of the flip-flop circuit is then brought out
through an output amplifier stage. The flip-flop circuit also operates a transistor inside the IC, the transistor collector usually being driven low to discharge a timing capacitor.
VCC
Threshold
6
8
R
Control
voltage
+
( 23 VCC)
5
R
–
1
Output
stage
F/F
( 13 VCC)
Discharge
–
R
Figure 17.16
timer IC.
Details of 555
Output
7
+
2
1
3
2
Trigger
input
4
Reset
Vref
Astable Operation
One popular application of the 555 timer IC is as an astable multivibrator or clock
circuit. The following analysis of the operation of the 555 as an astable circuit includes details of the different parts of the unit and how the various inputs and outputs are utilized. Figure 17.17 shows an astable circuit built using an external resistor and capacitor to set the timing interval of the output signal.
Figure 17.17 Astable
multivibrator using 555 IC.
732
Capacitor C charges toward VCC through external resistors RA and RB. Referring
to Fig. 17.17, the capacitor voltage rises until it goes above 2VCC/3. This voltage is
the threshold voltage at pin 6, which drives comparator 1 to trigger the flip-flop so
that the output at pin 3 goes low. In addition, the discharge transistor is driven on,
causing the output at pin 7 to discharge the capacitor through resistor RB. The capacitor voltage then decreases until it drops below the trigger level (VCC/3). The flipflop is triggered so that the output goes back high and the discharge transistor is turned
off, so that the capacitor can again charge through resistors RA and RB toward VCC.
Figure 17.18a shows the capacitor and output waveforms resulting from the astable
circuit. Calculation of the time intervals during which the output is high and low can
be made using the relations
Thigh 0.7(RA RB)C
(17.3)
Tlow 0.7RBC
(17.4)
T period Thigh Tlow
(17.5)
The total period is
The frequency of the astable circuit is then calculated using*
1
1.44
f T
(RA 2RB)C
Figure 17.18
(17.6)
Astable multivibrator for Example 17.1: (a) circuit; (b) waveforms.
*The period can be directly calculated from
T 0.693(RA 2RB)C 0.7(RA 2RB)C
and the frequency from
1.44
f (RA 2RB)C
17.4
Timer IC Unit Operation
733
EXAMPLE 17.1
Determine the frequency and draw the output waveform for the circuit of Fig. 17.18a.
Solution
Using Eqs. (17.3) through (17.6) yields
Thigh 0.7(RA RB)C 0.7(7.5 103 7.5 103)(0.1 106)
1.05 ms
Tlow 0.7RBC 0.7(7.5 103)(0.1 106) 0.525 ms
T Thigh Tlow 1.05 ms 0.525 ms 1.575 ms
1
1
f 635 Hz
T
1.575 103
The waveforms are drawn in Fig. 17.18b.
Monostable Operation
The 555 timer can also be used as a one-shot or monostable multivibrator circuit, as
shown in Fig. 17.19. When the trigger input signal goes negative, it triggers the oneshot, with output at pin 3 then going high for a time period
Thigh 1.1RAC
(17.7)
Referring back to Fig. 17.16, the negative edge of the trigger input causes comparator 2 to trigger the flip-flop, with the output at pin 3 going high. Capacitor C charges
toward VCC through resistor RA. During the charge interval, the output remains high.
When the voltage across the capacitor reaches the threshold level of 2VCC/3, comparator 1 triggers the flip-flop, with output going low. The discharge transistor also
goes low, causing the capacitor to remain at near 0 V until triggered again.
Figure 17.19b shows the input trigger signal and the resulting output waveform
for the 555 timer operated as a one-shot. Time periods for this circuit can range from
microseconds to many seconds, making this IC useful for a range of applications.
Figure 17.19
734
Chapter 17
Operation of 555 timer as one-shot: (a) circuit; (b) waveforms.
Linear-Digital ICs
Determine the period of the output waveform for the circuit of Fig. 17.20 when triggered by a negative pulse.
EXAMPLE 17.2
Figure 17.20 Monostable circuit for Example 17.2.
Solution
Using Eq. (17.7), we obtain
Thigh 1.1RAC 1.1(7.5 103)(0.1 106) 0.825 ms
17.5 VOLTAGE-CONTROLLED
OSCILLATOR
A voltage-controlled oscillator (VCO) is a circuit that provides a varying output signal (typically of square-wave or triangular-wave form) whose frequency can be adjusted over a range controlled by a dc voltage. An example of a VCO is the 566 IC
unit, which contains circuitry to generate both square-wave and triangular-wave signals whose frequency is set by an external resistor and capacitor and then varied by
an applied dc voltage. Figure 17.21a shows that the 566 contains current sources to
charge and discharge an external capacitor C1 at a rate set by external resistor R1 and
the modulating dc input voltage. A Schmitt trigger circuit is used to switch the current sources between charging and discharging the capacitor, and the triangular voltage developed across the capacitor and square wave from the Schmitt trigger are provided as outputs through buffer amplifiers.
Figure 17.21b shows the pin connection of the 566 unit and a summary of formula and value limitations. The oscillator can be programmed over a 10-to-1 frequency range by proper selection of an external resistor and capacitor, and then modulated over a 10-to-1 frequency range by a control voltage, VC.
A free-running or center-operating frequency, fo, can be calculated from
2
V VC
fo R1C1
V
(17.8)
with the following practical circuit value restrictions:
1. R1 should be within the range 2 k R1 20 k.
2. VC should be within range 34V VC V.
17.5
Voltage-Controlled Oscillator
735
Figure 17.21 A 566 function
generator: (a) block diagram;
(b) pin configuration and summary
of operating data.
3. fo should be below 1 MHz.
4. V should range between 10 V and 24 V.
Figure 17.22 shows an example in which the 566 function generator is used to
provide both square-wave and triangular-wave signals at a fixed frequency set by R1,
C1, and VC. A resistor divider R2 and R3 sets the dc modulating voltage at a fixed
value
Figure 17.22 Connection of
566 VCO unit.
736
Chapter 17
Linear-Digital ICs
R3
10 k
VC V (12 V) 10.4 V
R2 R3
1.5 k 10 k
(which falls properly in the voltage range 0.75V 9 V and V 12 V). Using Eq.
(17.8) yields
12 10.4
2
32.5 kHz
fo (10 103)(820 1012)
12
The circuit of Fig. 17.23 shows how the output square-wave frequency can be adjusted using the input voltage, VC, to vary the signal frequency. Potentiometer R3 allows varying VC from about 9 V to near 12 V, over the full 10-to-1 frequency range.
With the potentiometer wiper set at the top, the control voltage is
5 k 18 k
R3 R4
(V) (12 V) 11.74 V
VC 510
5 k 18 k
R2 R3 R4
resulting in a lower output frequency of
2
12 11.74
19.7 kHz
fo (10 103)(220 1012)
12
Figure 17.23 Connection of
566 as a VCO unit.
With the wiper arm of R3 set at the bottom, the control voltage is
18 k
R4
) (12 V) 9.19 V
VC (V
510 5 k 18 k
R2 R3 R4
resulting in an upper frequency of
2
12 9.19
212.9 kHz
fo (10 103)(220 1012)
12
The frequency of the output square wave can then be varied using potentiometer R3
over a frequency range of at least 10 to 1.
Rather than varying a potentiometer setting to change the value of VC, an input
modulating voltage, Vin, can be applied as shown in Fig. 17.24. The voltage divider
sets VC at about 10.4 V. An input ac voltage of about 1.4 V peak can drive VC around
the bias point between voltages of 9 and 11.8 V, causing the output frequency to vary
over about a 10-to-1 range. The input signal Vin thus frequency-modulates the output
voltage around the center frequency set by the bias value of VC 10.4 V ( fo 121.2 kHz).
17.5
Voltage-Controlled Oscillator
737
Figure 17.24 Operation of
VCO with frequency-modulating
input.
17.6 PHASE-LOCKED LOOP
A phase-locked loop (PLL) is an electronic circuit that consists of a phase detector,
a low-pass filter, and a voltage-controlled oscillator connected as shown in Fig. 17.25.
Common applications of a PLL include: (1) frequency synthesizers that provide multiples of a reference signal frequency [e.g., the carrier frequency for the multiple channels of a citizens’ band (CB) unit or marine-radio-band unit can be generated using
a single-crystal-controlled frequency and its multiples generated using a PLL]; (2)
FM demodulation networks for FM operation with excellent linearity between the input signal frequency and the PLL output voltage; (3) demodulation of the two data
transmission or carrier frequencies in digital-data transmission used in frequency-shift
keying (FSK) operation; and (4) a wide variety of areas including modems, telemetry receivers and transmitters, tone decoders, AM detectors, and tracking filters.
An input signal, Vi, and that from a VCO, Vo, are compared by a phase comparator
(refer to Fig. 17.25) providing an output voltage, Ve, that represents the phase difference between the two signals. This voltage is then fed to a low-pass filter that pro-
Figure 17.25
738
Chapter 17
Block diagram of basic phase-locked loop (PLL).
Linear-Digital ICs
vides an output voltage (amplified if necessary) that can be taken as the output voltage from the PLL and is used internally as the voltage to modulate the VCO’s frequency. The closed-loop operation of the circuit is to maintain the VCO frequency
locked to that of the input signal frequency.
Basic PLL Operation
The basic operation of a PLL circuit can be explained using the circuit of Fig. 17.25
as reference. We will first consider the operation of the various circuits in the phaselocked loop when the loop is operating in lock (the input signal frequency and the
VCO frequency are the same). When the input signal frequency is the same as that
from the VCO to the comparator, the voltage, Vd, taken as output is the value needed
to hold the VCO in lock with the input signal. The VCO then provides output of a
fixed-amplitude square-wave signal at the frequency of the input. Best operation is
obtained if the VCO center frequency, fo, is set with the dc bias voltage midway in
its linear operating range. The amplifier allows this adjustment in dc voltage from that
obtained as output of the filter circuit. When the loop is in lock, the two signals to
the comparator are of the same frequency, although not necessarily in phase. A fixed
phase difference between the two signals to the comparator results in a fixed dc voltage to the VCO. Changes in the input signal frequency then result in change in the
dc voltage to the VCO. Within a capture-and-lock frequency range, the dc voltage
will drive the VCO frequency to match that of the input.
While the loop is trying to achieve lock, the output of the phase comparator contains frequency components at the sum and difference of the signals compared. A lowpass filter passes only the lower-frequency component of the signal so that the loop
can obtain lock between input and VCO signals.
Owing to the limited operating range of the VCO and the feedback connection of
the PLL circuit, there are two important frequency bands specified for a PLL. The
capture range of a PLL is the frequency range centered about the VCO free-running
frequency, fo, over which the loop can acquire lock with the input signal. Once the
PLL has achieved capture, it can maintain lock with the input signal over a somewhat
wider frequency range called the lock range.
Applications
The PLL can be used in a wide variety of applications, including (1) frequency demodulation, (2) frequency synthesis, and (3) FSK decoders. Examples of each of these
follow.
FREQUENCY DEMODULATION
FM demodulation or detection can be directly achieved using the PLL circuit. If
the PLL center frequency is selected or designed at the FM carrier frequency, the filtered or output voltage of the circuit of Fig. 17.25 is the desired demodulated voltage, varying in value proportional to the variation of the signal frequency. The PLL
circuit thus operates as a complete intermediate-frequency (IF) strip, limiter, and demodulator as used in FM receivers.
One popular PLL unit is the 565, shown in Fig. 17.26a. The 565 contains a phase
detector, amplifier, and voltage-controlled oscillator, which are only partially connected internally. An external resistor and capacitor, R1 and C1, are used to set the
free-running or center frequency of the VCO. Another external capacitor, C2, is used
to set the low-pass filter passband, and the VCO output must be connected back as
input to the phase detector to close the PLL loop. The 565 typically uses two power
supplies, V and V.
17.6
Phase-Locked Loop
739
V+
10
565
C2
2
Input
3.6 kΩ
Phase
detector
3
Amp.
5
4
7
Demodulated
output
6
Reference
output
VCO
Output
9
8
R1
1
V−
C1
(a)
+6 V
10
C2
330 pF
Demodulated
output
2
FM signal
input
3.6 kΩ
Phase
detector
3
Amp.
7
5
Reference
output
6
4
VCO
Output
8
R1
10 kΩ
V7
+6 V
+5.3 V
9
C1
220 pF
1
fo =
0.3
R1 C1
−6 V
(b)
+5 V
+4.7 V
Frequency
fo
fL
fL
(= 136.36 kHz) fo +
2
2
(= 45.45 kHz)
(= 227.27 kHz)
± fL
(± 181.82 kHz)
fo −
Figure 17.26 Phase-locked loop
(PLL): (a) basic block diagram:
(b) PLL connected as a frequency
demodulator: (c) output voltage
vs. frequency plot.
(c)
Figure 17.26b shows the PLL connected to work as an FM demodulator. Resistor R1 and capacitor C1 set the free-running frequency, fo,
0.3
fo R1C1
(17.9)
0.3
136.36 kHz
3
(10 10 )(220 1012)
with limitation 2 k
R1
20 k. The lock range is
8f
fL o
V
8(136.36 103)
181.8 kHz
6
740
Chapter 17
Linear-Digital ICs
for supply voltages V 6 V. The capture range is
2 (181.8 10 )
1
156.1 kHz
(3.6 10 )(330 10 )
2 1
2 fL
fC 2
R2C2
3
12
3
The signal at pin 4 is a 136.36-kHz square wave. An input within the lock range of
181.8 kHz will result in the output at pin 7 varying around its dc voltage level set
with input signal at fo. Figure 17.26c shows the output at pin 7 as a function of the
input signal frequency. The dc voltage at pin 7 is linearly related to the input signal
frequency within the frequency range fL 181.8 kHz around the center frequency
136.36 kHz. The output voltage is the demodulated signal that varies with frequency
within the operating range specified.
FREQUENCY SYNTHESIS
A frequency synthesizer can be built around a PLL as shown in Fig. 17.27. A frequency divider is inserted between the VCO output and the phase comparator so that
the loop signal to the comparator is at frequency fo while the VCO output is Nfo. This
output is a multiple of the input frequency as long as the loop is in lock. The input
signal can be stabilized at f1 with the resulting VCO output at Nf1 if the loop is set
Input
f1
Phase
comparator
Low-pass
filter
Amp.
fo
÷N
VCO
Output
Nfo
(a)
+5 V
C2
330 pF
R1
10 kΩ
5.1 kΩ
8
7
10
4
2
14
Nfo
565
Vi
f1
Input A
10 kΩ
5
5
3
9
9
1
7490
C1
220 pF
−5 V
1 12
8
QA ÷ 2
QB ÷ 4
QC ÷ 8
QD ÷ 16
10
11
3
6
2
7
R0 (1) R0 (2) R 9 (1) R 9 (2)
fo
(b)
Figure 17.27
Frequency synthesizer: (a) block diagram; (b) implementation using 565 PLL unit.
17.6
Phase-Locked Loop
741
up to lock at the fundamental frequency (when fo f1). Figure 17.27b shows an example using a 565 PLL as frequency multiplier and a 7490 as divider. The input Vi
at frequency f1 is compared to the input (frequency fo) at pin 5. An output at Nfo (4fo
in the present example) is connected through an inverter circuit to provide an input
at pin 14 of the 7490, which varies between 0 and 5 V. Using the output at pin 9,
which is divided by 4 from that at the input to the 7490, the signal at pin 4 of the
PLL is four times the input frequency as long as the loop remains in lock. Since the
VCO can vary over only a limited range from its center frequency, it may be necessary to change the VCO frequency whenever the divider value is changed. As long as
the PLL circuit is in lock, the VCO output frequency will be exactly N times the input frequency. It is only necessary to readjust fo to be within the capture-and-lock
range, the closed loop then resulting in the VCO output becoming exactly Nf1 at lock.
FSK DECODERS
An FSK (frequency-shift keyed) signal decoder can be built as shown in Fig. 17.28.
The decoder receives a signal at one of two distinct carrier frequencies, 1270 Hz or
1070 Hz, representing the RS-232C logic levels or mark (5 V) or space (14 V),
respectively. As the signal appears at the input, the loop locks to the input frequency
and tracks it between two possible frequencies with a corresponding dc shift at the
output.
Figure 17.28 Connection of
565 as FSK decoder.
The RC ladder filter (three sections of C 0.02 F and R 10 k) is used to
remove the sum frequency component. The free-running frequency is adjusted with
R1 so that the dc voltage level at the output (pin 7) is the same as that at pin 6. Then
an input at frequency 1070 Hz will drive the decoder output voltage to a more positive voltage level, driving the digital output to the high level (space or 14 V). An
input at 1270 Hz will correspondingly drive the 565 dc output less positive with the
digital output, which then drops to the low level (mark or 5 V).
17.7 INTERFACING CIRCUITRY
Connecting different types of circuits, either in digital or analog circuits, may require
some sort of interfacing circuit. An interface circuit may be used to drive a load or to
obtain a signal as a receiver circuit. A driver circuit provides the output signal at a
742
Chapter 17
Linear-Digital ICs
voltage or current level suitable to operate a number of loads, or to operate such devices as relays, displays, or power units. A receiver circuit essentially accepts an input signal, providing high input impedance to minimize loading of the input signal.
Furthermore, the interface circuits may include strobing, which provides connecting
the interface signals during specific time intervals established by the strobe.
Figure 17.29a shows a dual-line driver, each driver accepting input of TTL signals, providing output capable of driving TTL or MOS device circuits. This type of
interface circuit comes in various forms, some as inverting and others as noninverting units. The circuit of Fig. 17.29b shows a dual-line receiver having both inverting
and noninverting inputs so that either operating condition can be selected. As an example, connection of an input signal to the inverting input would result in an inverted
output from the receiver unit. Connecting the input to the noninverting input would
provide the same interfacing except that the output obtained would have the same polarity as the received signal. The driver-receiver unit of Fig. 17.29 provides an output
when the strobe signal is present (high in this case).
Figure 17.29 Interface units: (a) dual-line drivers (SN75150); (b) dual-line
receivers (SN75152).
Another type of interface circuit is that used to connect various digital input and
output units, signals with devices such as keyboards, video terminals, and printers.
One of the EIA electronic industry standards is referred to as RS-232C. This standard
states that a digital signal represents a mark (logic-1) and a space (logic-0). The definitions of mark and space vary with the type of circuit used (although a full reading
of the standard will spell out the acceptable limits of mark and space signals).
17.7 Interfacing Circuitry
743
RS-232C-to-TTL Converter
For TTL circuits, 5 V is a mark and 0 V is a space. For RS-232C, a mark could be
12 V and a space 12 V. Figure 17.30a provides a tabulation of some mark and
space definitions. For a unit having outputs defined by RS-232C that is to operate
into another unit operating with a TTL signal level, an interface circuit as shown in
Fig. 17.30b could be used. A mark output from the driver (at 12 V) would be clipped
by the diode so that the input to the inverter circuit is near 0 V, resulting in an output of 5 V (TTL mark). A space output at 12 V would drive the inverter output
low for a 0-V output (a space).
Figure 17.30
Interfacing signal standards and converter circuits.
Another example of an interface circuit converts the signals from a TTY current
loop into TTL levels as shown in Fig. 17.30c. An input mark results when 20 mA of
current is drawn from the source through the output line of the teletype (TTY). This
current then goes through the diode element of an opto-isolator, driving the output
transistor on. The input to the inverter going low results in a 5-V signal from the
7407 inverter output so that a mark from the teletype results in a mark to the TTL
input. A space from the teletype current loop provides no current, with the optoisolator transistor remaining off and the inverter output then 0 V, which is a TTL
space signal.
Another means of interfacing digital signals is made using open-collector output
or tri-state output. When a signal is output from a transistor collector (see Fig. 17.31)
that is not connected to any other electronic component, the output is open-collector.
This permits connecting a number of signals to the same wire or bus. Any transistor
going on then provides a low output voltage, while all transistors remaining off provide a high output voltage.
744
Chapter 17
Linear-Digital ICs
Figure 17.31
Connections to data lines: (a) open-collector output; (b) tri-state output.
17.8 PSPICE WINDOWS
Many of the practical op-amp applications covered in this chapter can be analyzed
using PSpice. Analysis of various problems can display the resulting dc bias, or one
can use PROBE to display resulting waveforms.
Program 17.1—Comparator Circuit Used to Drive an LED
Using Design Center, draw the circuit of a comparator circuit with output driving an
LED indicator as shown in Fig. 17.32. To be able to view the magnitude of the dc
output voltage, place a VPRINT1 component at Vo with DC and MAG selected. To
view the dc current through the LED, place an IPRINT component in series with the
LED current meter as shown in Fig. 17.32. The Analysis Setup provides for a dc
sweep as shown in Fig. 17.33. The DC Sweep is set, as shown, for Vi from 4 to 8 V
in 1-V steps. After running the simultation, some of the resulting analysis output obtained is shown in Fig. 17.34.
Figure 17.32
Comparator circuit used to drive an LED.
17.8
PSpice Windows
745
Figure 17.33
Analysis Setup for a dc sweep of the circuit of Fig. 17.32.
The circuit of Fig. 17.32 shows a voltage divider which provides 6 V to the minus input so that any input (Vi) below 6 V will result in the output at the minus saturation voltage (near 10 V). Any input above 6 V results in the output going to
the positive saturation level (near 10 V). The LED will therefore be driven on by
any input above the reference level of 6 V and left off by any input below 6 V.
The listing of Fig. 17.34 shows a table of the output voltage and a table of the LED
current for inputs from 4 to 8 V. The table shows that the LED current is nearly 0 for
inputs up to 6 V and that a current of about 20 mA lights the LED for inputs at 6
V or above.
Program 17.2—Comparator Operation
Figure 17.34 Analysis output
(edited) for circuit of Fig. 17.32.
The operation of a comparator IC can be demonstrated using a 741 op-amp as shown
in Fig. 17.35. The input is a 5 V, peak sinusoidal signa. The Analysis Setup provides
for Transient analysis with Print Step of 20 ns and Final Time of 3 ms. Since the
input signal is applied to the noninverting input, the output is in-phase with the input. When the input goes above 0 V, the output goes to the positive saturation level,
746
Chapter 17
Linear-Digital ICs
Figure 17.35
comparator.
Schematic for a
near 5 V. When the input goes below 0 V, the output goes to the negative saturation level—this being 0 V since the minus voltage input is set to that value. Figure
17.36 shows a PROBE output of input and output voltages.
Figure 17.36
Probe output for the comparator of Fig. 17.35.
Program 17.3—Operation of 555 Timer as Oscillator
Figure 17.37 shows a 555 timer connected as an oscillator. Equations (17.3) and (17.4)
can be used to calculate the charge and discharge times as follows:
Thigh 0.7(RA RB)C 0.7(7.5 k 7.15 k)(0.1 F) 1.05 ms
Tlow 0.7RBC 0.7(7.5 k)(0.1 F) 0.525 ms
Figure 17.37
Schematic of a 555 timer oscillator.
17.8
PSpice Windows
747
Figure 17.38
Probe output for the 555 oscillator of Fig. 17.37.
The resulting trigger and output waveforms are shown in Fig. 17.38. When the trigger charges to the upper trigger level, the output goes to the low output level of 0 V.
The output stays low until the trigger input discharges to the low trigger level, at which
time the output goes to the high level of 5 V.
PROBLEMS
§ 17.2 Comparator Unit Operation
1. Draw the diagram of a 741 op-amp operated from 15-V supplies with Vi() 0 V and
Vi() 5 V. Include terminal pin connections.
2. Sketch the output waveform for the circuit of Fig. 17.39.
3. Draw a circuit diagram of a 311 op-amp showing an input of 10 V rms applied to the inverting input and the plus input to ground. Identify all pin numbers.
4. Draw the resulting output waveform for the circuit of Fig. 17.40.
Figure 17.39
Problem 2
Figure 17.40
Problem 4
5. Draw the circuit diagram of a zero-crossing detector using a 339 comparator stage with 12-V
supplies.
748
Chapter 17
Linear-Digital ICs
6. Sketch the output waveform for the circuit of Fig. 17.41.
Figure 17.41
*
Problem 6
7. Describe the operation of the circuit in Fig. 17.42.
Figure 17.42
Problem 7
§ 17.3 Digital–Analog Converters
8. Sketch a five-stage ladder network using 15-k and 30-k resistors.
9. For a reference voltage of 16 V, calculate the output voltage for an input of 11010 to the circuit of Problem 8.
10. What voltage resolution is possible using a 12-stage ladder network with a 10-V reference voltage?
Problems
749
11. For a dual-slope converter, describe what occurs during the fixed time interval and the count
interval.
12. How many count steps occur using a 12-stage digital counter at the output of an ADC?
13. What is the maximum count interval using a 12-stage counter operated at a clock rate of
20 MHz?
§ 17.4 Timer IC Unit Operation
14. Sketch the circuit of a 555 timer connected as an astable multivibrator for operation at
350 kHz. Determine the value of capacitor, C, needed using RA RB 7.5 k.
15. Draw the circuit of a one-shot using a 555 timer to provide one time period of 20 s. If RA 7.5 k, what value of C is needed?
16. Sketch the input and output waveforms for a one-shot using a 555 timer triggered by a 10-kHz
clock for RA 5.1 k and C 5 nF.
§ 17.5 Voltage-Controlled Oscillator
17. Calculate the center frequency of a VCO using a 566 IC as in Fig. 17.22 for R1 4.7 k,
R2 1.8 k, R3 11 k, and C1 0.001 F.
*
18. What frequency range results in the circuit of Fig. 17.23 for C1 0.001 F?
19. Determine the capacitor needed in the circuit of Fig. 17.22 to obtain a 200-kHz output.
§ 17.6 Phase-Locked Loop
20. Calculate the VCO free-running frequency for the circuit of Fig. 17.26b with R1 4.7 k and
C1 0.001 F.
21. What value of capacitor, C1, is required in the circuit of Fig. 17.26b to obtain a center frequency of 100 kHz?
22. What is the lock range of the PLL circuit in Fig. 17.26b for R1 4.7 k and C1 0.001 F?
§ 17.7 Interfacing Circuitry
23. Describe the signal conditions for current-loop and RS-232C interfaces.
24. What is a data bus?
25. What is the difference between open-collector and tri-state output?
§ 17.8 PSpice Windows
*
26. Use Design Center to draw a schematic circuit as in Fig. 17.32, using an LM111 with Vi 5 V
rms applied to minus () input and 5 V rms applied to plus () input. Use Probe to view
the output waveform.
*
27. Use Design Center to draw a schematic circuit as in Fig. 17.35. Examine the output listing for
the results.
*
28. Use Design Center to draw a 555 oscillator with resulting output with tlow 2 ms, thigh 5 ms.
*Please note: Asterisks indicate more difficult problems.
750
Chapter 17
Linear-Digital ICs
13
Sinusoidal Alternating
Waveforms
13.1
INTRODUCTION
The analysis thus far has been limited to dc networks, networks in
which the currents or voltages are fixed in magnitude except for transient effects. We will now turn our attention to the analysis of networks
in which the magnitude of the source varies in a set manner. Of particular interest is the time-varying voltage that is commercially available
in large quantities and is commonly called the ac voltage. (The letters
ac are an abbreviation for alternating current.) To be absolutely rigorous, the terminology ac voltage or ac current is not sufficient to
describe the type of signal we will be analyzing. Each waveform of Fig.
13.1 is an alternating waveform available from commercial supplies.
The term alternating indicates only that the waveform alternates
between two prescribed levels in a set time sequence (Fig. 13.1). To be
v
v
0
t
Sinusoidal
0
v
t 0
Square wave
t
Triangular wave
FIG. 13.1
Alternating waveforms.
absolutely correct, the term sinusoidal, square wave, or triangular must
also be applied. The pattern of particular interest is the sinusoidal ac
waveform for voltage of Fig. 13.1. Since this type of signal is encountered in the vast majority of instances, the abbreviated phrases ac voltage and ac current are commonly applied without confusion. For the
other patterns of Fig. 13.1, the descriptive term is always present, but
frequently the ac abbreviation is dropped, resulting in the designation
square-wave or triangular waveforms.
510

SINUSOIDAL ALTERNATING WAVEFORMS
One of the important reasons for concentrating on the sinusoidal ac
voltage is that it is the voltage generated by utilities throughout the
world. Other reasons include its application throughout electrical, electronic, communication, and industrial systems. In addition, the chapters
to follow will reveal that the waveform itself has a number of characteristics that will result in a unique response when it is applied to the
basic electrical elements. The wide range of theorems and methods
introduced for dc networks will also be applied to sinusoidal ac systems. Although the application of sinusoidal signals will raise the
required math level, once the notation given in Chapter 14 is understood, most of the concepts introduced in the dc chapters can be applied
to ac networks with a minimum of added difficulty.
The increasing number of computer systems used in the industrial
community requires, at the very least, a brief introduction to the terminology employed with pulse waveforms and the response of some fundamental configurations to the application of such signals. Chapter 22
will serve such a purpose.
13.2 SINUSOIDAL ac VOLTAGE
CHARACTERISTICS AND DEFINITIONS
Generation
Sinusoidal ac voltages are available from a variety of sources. The
most common source is the typical home outlet, which provides an ac
voltage that originates at a power plant; such a power plant is most
commonly fueled by water power, oil, gas, or nuclear fusion. In each
case an ac generator (also called an alternator), as shown in Fig.
13.2(a), is the primary component in the energy-conversion process.
Inverter
(a)
(b)
(c)
(d)
(e)
FIG. 13.2
Various sources of ac power: (a) generating plant; (b) portable ac generator;
(c) wind-power station; (d) solar panel; (e) function generator.
The power to the shaft developed by one of the energy sources listed
will turn a rotor (constructed of alternating magnetic poles) inside a
set of windings housed in the stator (the stationary part of the
dynamo) and will induce a voltage across the windings of the stator,
as defined by Faraday’s law,
SINUSOIDAL ac VOLTAGE CHARACTERISTICS AND DEFINITIONS
df
eN dt
Through proper design of the generator, a sinusoidal ac voltage is
developed that can be transformed to higher levels for distribution
through the power lines to the consumer. For isolated locations where
power lines have not been installed, portable ac generators [Fig.
13.2(b)] are available that run on gasoline. As in the larger power
plants, however, an ac generator is an integral part of the design.
In an effort to conserve our natural resources, wind power and solar
energy are receiving increasing interest from various districts of the world
that have such energy sources available in level and duration that make the
conversion process viable. The turning propellers of the wind-power station [Fig. 13.2(c)] are connected directly to the shaft of an ac generator to
provide the ac voltage described above. Through light energy absorbed in
the form of photons, solar cells [Fig. 13.2(d)] can generate dc voltages.
Through an electronic package called an inverter, the dc voltage can be
converted to one of a sinusoidal nature. Boats, recreational vehicles (RVs),
etc., make frequent use of the inversion process in isolated areas.
Sinusoidal ac voltages with characteristics that can be controlled by
the user are available from function generators, such as the one in Fig.
13.2(e). By setting the various switches and controlling the position of
the knobs on the face of the instrument, one can make available sinusoidal voltages of different peak values and different repetition rates.
The function generator plays an integral role in the investigation of the
variety of theorems, methods of analysis, and topics to be introduced in
the chapters that follow.
Definitions
The sinusoidal waveform of Fig. 13.3 with its additional notation will now
be used as a model in defining a few basic terms. These terms can, howe
T1
T2
Max
e1
0
Em
t2
t1
Ep –p
e2
Em
Max
t
T3
FIG. 13.3
Important parameters for a sinusoidal voltage.
ever, be applied to any alternating waveform. It is important to remember
as you proceed through the various definitions that the vertical scaling is
in volts or amperes and the horizontal scaling is always in units of time.
Waveform: The path traced by a quantity, such as the voltage in
Fig. 13.3, plotted as a function of some variable such as time (as
above), position, degrees, radians, temperature, and so on.

511
512

SINUSOIDAL ALTERNATING WAVEFORMS
Instantaneous value: The magnitude of a waveform at any instant
of time; denoted by lowercase letters (e1, e2).
Peak amplitude: The maximum value of a waveform as measured
from its average, or mean, value, denoted by uppercase letters (such
as Em for sources of voltage and Vm for the voltage drop across a
load). For the waveform of Fig. 13.3, the average value is zero volts
and Em is as defined by the figure.
Peak value: The maximum instantaneous value of a function as
measured from the zero-volt level. For the waveform of Fig. 13.3,
the peak amplitude and peak value are the same, since the average
value of the function is zero volts.
Peak-to-peak value: Denoted by Ep-p or Vp-p, the full voltage
between positive and negative peaks of the waveform, that is, the
sum of the magnitude of the positive and negative peaks.
Periodic waveform: A waveform that continually repeats itself
after the same time interval. The waveform of Fig. 13.3 is a periodic
waveform.
Period (T ): The time interval between successive repetitions of a
periodic waveform (the period T1 T2 T3 in Fig. 13.3), as long as
successive similar points of the periodic waveform are used in determining T.
Cycle: The portion of a waveform contained in one period of time.
The cycles within T1, T2, and T3 of Fig. 13.3 may appear different in
Fig. 13.4, but they are all bounded by one period of time and therefore satisfy the definition of a cycle.
FIG. 13.4
Defining the cycle and period of a sinusoidal waveform.
Frequency ( f ): The number of cycles that occur in 1 s. The frequency of the waveform of Fig. 13.5(a) is 1 cycle per second, and
for Fig. 13.5(b), 21⁄2 cycles per second. If a waveform of similar
shape had a period of 0.5 s [Fig. 13.5(c)], the frequency would be 2
cycles per second.
FIG. 13.5
Demonstrating the effect of a changing frequency on the period of a sinusoidal
waveform.
SINUSOIDAL ac VOLTAGE CHARACTERISTICS AND DEFINITIONS

513
The unit of measure for frequency is the hertz (Hz), where
1 hertz (Hz) 1 cycle per second (c/s)
(13.1)
The unit hertz is derived from the surname of Heinrich Rudolph Hertz
(Fig. 13.6), who did original research in the area of alternating currents
and voltages and their effect on the basic R, L, and C elements. The frequency standard for North America is 60 Hz, whereas for Europe it is
predominantly 50 Hz.
As with all standards, any variation from the norm will cause difficulties. In 1993, Berlin, Germany, received all its power from eastern plants, whose output frequency was varying between 50.03 and
51 Hz. The result was that clocks were gaining as much as 4 minutes a day. Alarms went off too soon, VCRs clicked off before the
end of the program, etc., requiring that clocks be continually reset. In
1994, however, when power was linked with the rest of Europe, the
precise standard of 50 Hz was reestablished and everyone was on
time again.
Using a log scale (described in detail in Chapter 21), a frequency
spectrum from 1 GHz to 1000 GHz can be scaled off on the same
axis, as shown in Fig. 13.7. A number of terms in the various spectrums are probably familiar to the reader from everyday experiences.
Note that the audio range (human ear) extends from only 15 Hz to 20
kHz, but the transmission of radio signals can occur between 3 kHz
and 300 GHz. The uniform process of defining the intervals of the
radio-frequency spectrum from VLF to EHF is quite evident from the
length of the bars in the figure (although keep in mind that it is a log
scale, so the frequencies encompassed within each segment are quite
different). Other frequencies of particular interest (TV, CB, microwave, etc.) are also included for reference purposes. Although it is
numerically easy to talk about frequencies in the megahertz and gigahertz range, keep in mind that a frequency of 100 MHz, for instance,
represents a sinusoidal waveform that passes through 100,000,000
cycles in only 1 s—an incredible number when we compare it to the
60 Hz of our conventional power sources The new Pentium II chip
manufactured by Intel can run at speeds up to 450 MHz. Imagine a
product able to handle 450,000,000 instructions per second—an
incredible achievement.
Since the frequency is inversely related to the period—that is, as one
increases, the other decreases by an equal amount—the two can be
related by the following equation:
1
f T
or
f Hz
T seconds (s)
1
T f
(13.2)
(13.3)
German (Hamburg,
Berlin, Karlsruhe)
(1857–94)
Physicist
Professor of Physics,
Karlsruhe
Polytechnic and
University of Bonn
Courtesy of the
Smithsonian Institution
Photo No. 66,606
Spurred on by the earlier predictions of the English
physicist James Clerk Maxwell, Heinrich Hertz produced electromagnetic waves in his laboratory at the
Karlsruhe Polytechnic while in his early 30s. The
rudimentary transmitter and receiver were in essence the first to broadcast and receive radio waves.
He was able to measure the wavelength of the
electromagnetic waves and confirmed that the velocity of propagation is in the same order of magnitude as light. In addition, he demonstrated that the
reflective and refractive properties of electromagnetic waves are the same as those for heat and light
waves. It was indeed unfortunate that such an ingenious, industrious individual should pass away at the
very early age of 37 due to a bone disease.
FIG. 13.6
Heinrich Rudolph Hertz.
514

SINUSOIDAL ALTERNATING WAVEFORMS
Microwave
Microwave
oven
EHF
SHF
30 GHz – 300 GHz
(Extremely High Freq.)
3 GHz – 30 GHz (Super-High Freq.)
UHF
VHF
HF
MF
LF
VLF
300 MHz – 3 GHz (Ultrahigh Freq.)
30 MHz – 300 MHz (Very High Freq.)
3 MHz – 30 MHz (High Freq.)
300 kHz – 3 MHz (Medium Freq.)
30 kHz – 300 kHz (Low Freq.)
3 kHz – 30 kHz (Very Low Freq.)
Radio frequencies (spectrum)
Infrared
3 kHz – 300 GHz
Audio frequencies
15 Hz – 20 kHz
1 Hz
10 Hz
100 Hz
1 kHz
10 kHz 100 kHz 1 MHz 10 MHz 100 MHz 1 GHz 10 GHz 100 GHz 1000 GHz f (log scale)
FM
88 MHz – 108 MHz
TV channels (2 – 6)
54 MHz – 88 MHz
TV channels (7 – 13)
174 MHz – 216 MHz
TV channels (14 – 83)
470 MHz – 890 MHz
CB
Countertop microwave oven
26.9 MHz – 27.4 MHz
2.45 GHz
Shortwave
1.5 MHz – 30 MHz
Cordless telephones
46 MHz – 49 MHz
Pagers VHF
30 MHz – 50 MHz
Pagers UHF
405 MHz – 512 MHz
FIG. 13.7
Areas of application for specific frequency bands.
SINUSOIDAL ac VOLTAGE CHARACTERISTICS AND DEFINITIONS

515
EXAMPLE 13.1 Find the period of a periodic waveform with a frequency of
a. 60 Hz.
b. 1000 Hz.
Solutions:
1
1
a. T 0.01667 s or 16.67 ms
f
60 Hz
(a recurring value since 60 Hz is so prevalent)
e
10 V
1
1
b. T 103 s 1 ms
f
1000 Hz
0
5
15
35
25
t (ms)
EXAMPLE 13.2 Determine the frequency of the waveform of Fig.
13.8.
Solution:
From the figure, T (25 ms 5 ms) 20 ms, and
FIG. 13.8
Example 13.2.
1
1
f 50 Hz
T
20 103 s
EXAMPLE 13.3 The oscilloscope is an instrument that will display
alternating waveforms such as those described above. A sinusoidal pattern
appears on the oscilloscope of Fig. 13.9 with the indicated vertical and
horizontal sensitivities. The vertical sensitivity defines the voltage associated with each vertical division of the display. Virtually all oscilloscope
screens are cut into a crosshatch pattern of lines separated by 1 cm in the
vertical and horizontal directions. The horizontal sensitivity defines the
time period associated with each horizontal division of the display.
For the pattern of Fig. 13.9 and the indicated sensitivities, determine
the period, frequency, and peak value of the waveform.
Solution:
Vertical sensitivity = 0.1 V/div.
Horizontal sensitivity = 50 s/div.
FIG. 13.9
Example 13.3.
One cycle spans 4 divisions. The period is therefore
50 ms
T 4 div. 200 ms
div.
and the frequency is
1
1
f 5 kHz
200 106 s
T
The vertical height above the horizontal axis encompasses 2 divisions.
Therefore,
0.1 V
Vm 2 div. 0.2 V
div.
e
Defined Polarities and Direction
In the following analysis, we will find it necessary to establish a set of
polarities for the sinusoidal ac voltage and a direction for the sinusoidal
ac current. In each case, the polarity and current direction will be for an
instant of time in the positive portion of the sinusoidal waveform. This
is shown in Fig. 13.10 with the symbols for the sinusoidal ac voltage
and current. A lowercase letter is employed for each to indicate that the
quantity is time dependent; that is, its magnitude will change with time.
i
+
t
t
e
i
–
(a)
(b)
FIG. 13.10
(a) Sinusoidal ac voltage sources;
(b) sinusoidal current sources.
516

SINUSOIDAL ALTERNATING WAVEFORMS
The need for defining polarities and current direction will become quite
obvious when we consider multisource ac networks. Note in the last
sentence the absence of the term sinusoidal before the phrase ac networks. This phrase will be used to an increasing degree as we progress;
sinusoidal is to be understood unless otherwise indicated.
i
t
+
R, L, or C
13.3
v
t
–
FIG. 13.11
The sine wave is the only alternating
waveform whose shape is not altered by the
response characteristics of a pure resistor,
inductor, or capacitor.
THE SINE WAVE
The terms defined in the previous section can be applied to any type of
periodic waveform, whether smooth or discontinuous. The sinusoidal
waveform is of particular importance, however, since it lends itself
readily to the mathematics and the physical phenomena associated with
electric circuits. Consider the power of the following statement:
The sine wave is the only alternating waveform whose shape is
unaffected by the response characteristics of R, L, and C elements.
In other words, if the voltage across (or current through) a resistor,
coil, or capacitor is sinusoidal in nature, the resulting current (or voltage, respectively) for each will also have sinusoidal characteristics, as
shown in Fig. 13.11. If a square wave or a triangular wave were
applied, such would not be the case. It must be pointed out that the
above statement is also applicable to the cosine wave, since the waves
differ only by a 90° shift on the horizontal axis, as shown in Fig. 13.12.
The unit of measurement for the horizontal axis of Fig. 13.12 is the
degree. A second unit of measurement frequently used is the radian
(rad). It is defined by a quadrant of a circle such as in Fig. 13.13 where
the distance subtended on the circumference equals the radius of the
circle.
If we define x as the number of intervals of r (the radius) around the
circumference of the circle, then
C 2pr x ⋅ r
and we find
x 2p
Therefore, there are 2p rad around a 360° circle, as shown in Fig.
13.14, and
FIG. 13.12
Sine wave and cosine wave with the
horizontal axis in degrees.
2p rad 360°
r
1 radian
57.296°
r
FIG. 13.13
Defining the radian.
FIG. 13.14
There are 2p radians in one full circle of 360°.
(13.4)
THE SINE WAVE
1 rad 57.296° 57.3°
with

517
(13.5)
A number of electrical formulas contain a multiplier of p. For this
reason, it is sometimes preferable to measure angles in radians rather
than in degrees.
The quantity p is the ratio of the circumference of a circle to its
diameter.
p has been determined to an extended number of places primarily in
an attempt to see if a repetitive sequence of numbers appears. It does
not. A sampling of the effort appears below:
p 3.14159 26535 89793 23846 26433 . . .
Although the approximation p 3.14 is often applied, all the calculations in this text will use the p function as provided on all scientific calculators.
For 180° and 360°, the two units of measurement are related as
shown in Fig. 13.14. The conversion equations between the two are the
following:
(13.6)
(13.7)
p
Radians (degrees)
180°
180°
Degrees (radians)
p
Applying these equations, we find
p
p
90°: Radians (90°) rad
2
180°
p
p
30°: Radians (30°) rad
6
180°
p
rad:
3
3p
rad:
2
180°
Degrees p
180°
Degrees p
3 60°
3p
2 270°
p
Using the radian as the unit of measurement for the abscissa, we would
obtain a sine wave, as shown in Fig. 13.15.
It is of particular interest that the sinusoidal waveform can be
derived from the length of the vertical projection of a radius vector
rotating in a uniform circular motion about a fixed point. Starting as
shown in Fig. 13.16(a) and plotting the amplitude (above and below
zero) on the coordinates drawn to the right [Figs. 13.16(b) through (i)],
we will trace a complete sinusoidal waveform after the radius vector
has completed a 360° rotation about the center.
The velocity with which the radius vector rotates about the center,
called the angular velocity, can be determined from the following
equation:
distance (degrees or radians)
Angular velocity time (seconds)
(13.8)
v, i, etc.
Sine wave
0
4
2
3
4
5 3 7
4
2 4 2
(rad)
FIG. 13.15
Plotting a sine wave versus radians.
518

SINUSOIDAL ALTERNATING WAVEFORMS
(a)
= 0° 0°
Note equality.
(b)
= 45°
0° 45°
(c)
(d)
= 90°
0°
90°
= 135°
0° 45° 90° 135°
(e)
= 180°
0°
180°
= 225°
225°
(f)
0°
= 270°
270°
(g)
0°
= 315°
(h)
315°
0°
Sine wave
= 360°
(i)
225° 270° 315° 360°
0° 45° 90° 135° 180°
T (period)
FIG. 13.16
Generating a sinusoidal waveform through the vertical projection of a
rotating vector.
THE SINE WAVE

Substituting into Eq. (13.8) and assigning the Greek letter omega (q)
to the angular velocity, we have
and
a
q t
(13.9)
a qt
(13.10)
Since q is typically provided in radians per second, the angle a
obtained using Eq. (13.10) is usually in radians. If a is required in
degrees, Equation (13.7) must be applied. The importance of remembering the above will become obvious in the examples to follow.
In Fig. 13.16, the time required to complete one revolution is equal
to the period (T) of the sinusoidal waveform of Fig. 13.16(i). The radians subtended in this time interval are 2p. Substituting, we have
2p
q T
(rad/s)
(13.11)
In words, this equation states that the smaller the period of the
sinusoidal waveform of Fig. 13.16(i), or the smaller the time interval
before one complete cycle is generated, the greater must be the angular velocity of the rotating radius vector. Certainly this statement
agrees with what we have learned thus far. We can now go one step
further and apply the fact that the frequency of the generated waveform is inversely related to the period of the waveform; that is, f 1/T. Thus,
q 2pf
(rad/s)
(13.12)
This equation states that the higher the frequency of the generated
sinusoidal waveform, the higher must be the angular velocity. Equations
(13.11) and (13.12) are verified somewhat by Fig. 13.17, where for the
same radius vector, q 100 rad/s and 500 rad/s.
EXAMPLE 13.4 Determine the angular velocity of a sine wave having a frequency of 60 Hz.
Solution:
q 2pf (2p)(60 Hz) 377 rad/s
(a recurring value due to 60-Hz predominance)
EXAMPLE 13.5 Determine the frequency and period of the sine wave
of Fig. 13.17(b).
Solution:
Since q 2p/T,
2p
2p rad
2p rad
T 12.57 ms
500 rad/s
500 rad/s
q
and
1
1
f 79.58 Hz
T
12.57 103 s
FIG. 13.17
Demonstrating the effect of q on the
frequency and period.
519
520

SINUSOIDAL ALTERNATING WAVEFORMS
EXAMPLE 13.6 Given q 200 rad/s, determine how long it will take
the sinusoidal waveform to pass through an angle of 90°.
Solution:
Eq. (13.10): a qt, and
a
t
q
However, a must be substituted as p/2 ( 90°) since q is in radians per
second:
p
a
p/2 rad
t 7.85 ms
200 rad/s
400 s
q
EXAMPLE 13.7 Find the angle through which a sinusoidal waveform
of 60 Hz will pass in a period of 5 ms.
Solution:
Eq. (13.11): a qt, or
a 2pft (2p)(60 Hz)(5 103 s) 1.885 rad
If not careful, one might be tempted to interpret the answer as
1.885°. However,
180°
a (°) (1.885 rad) 108°
p rad
13.4 GENERAL FORMAT FOR THE SINUSOIDAL
VOLTAGE OR CURRENT
The basic mathematical format for the sinusoidal waveform is
Am sin a
(13.13)
where Am is the peak value of the waveform and a is the unit of measure for the horizontal axis, as shown in Fig. 13.18.
FIG. 13.18
Basic sinusoidal function.
The equation a qt states that the angle a through which the rotating vector of Fig. 13.16 will pass is determined by the angular velocity
of the rotating vector and the length of time the vector rotates. For
example, for a particular angular velocity (fixed q), the longer the
radius vector is permitted to rotate (that is, the greater the value of t),
the greater will be the number of degrees or radians through which the
vector will pass. Relating this statement to the sinusoidal waveform, for
a particular angular velocity, the longer the time, the greater the num-
GENERAL FORMAT FOR THE SINUSOIDAL VOLTAGE OR CURRENT
ber of cycles shown. For a fixed time interval, the greater the angular
velocity, the greater the number of cycles generated.
Due to Eq. (13.10), the general format of a sine wave can also be
written
Am sin qt
(13.14)
with qt as the horizontal unit of measure.
For electrical quantities such as current and voltage, the general format is
i Im sin qt Im sin a
e Em sin qt Em sin a
where the capital letters with the subscript m represent the amplitude,
and the lowercase letters i and e represent the instantaneous value of
current or voltage, respectively, at any time t. This format is particularly
important since it presents the sinusoidal voltage or current as a function of time, which is the horizontal scale for the oscilloscope. Recall
that the horizontal sensitivity of a scope is in time per division and not
degrees per centimeter.
EXAMPLE 13.8 Given e 5 sin a, determine e at a 40° and a 0.8p .
Solution:
For a 40°,
e 5 sin 40° 5(0.6428) 3.214 V
For a 0.8p,
180°
a (°) (0.8p) 144°
p
and
e 5 sin 144° 5(0.5878) 2.939 V
The conversion to degrees will not be required for most modern-day
scientific calculators since they can perform the function directly. First, be
sure that the calculator is in the RAD mode. Then simply enter the radian
measure and use the appropriate trigonometric key (sin, cos, tan, etc.).
The angle at which a particular voltage level is attained can be
determined by rearranging the equation
e Em sin a
in the following manner:
e
sin a Em
which can be written
e
a sin1 Em
(13.15)
Similarly, for a particular current level,
i
a sin1 Im
The function sin1 is available on all scientific calculators.
(13.16)

521
522

SINUSOIDAL ALTERNATING WAVEFORMS
EXAMPLE 13.9
a. Determine the angle at which the magnitude of the sinusoidal function v 10 sin 377t is 4 V.
b. Determine the time at which the magnitude is attained.
Solutions:
a. Eq. (13.15):
v
4V
a1 sin1 sin1 sin1 0.4 23.578°
Em
10 V
However, Figure 13.19 reveals that the magnitude of 4 V (positive) will be attained at two points between 0° and 180°. The second
intersection is determined by
v (V)
a2 180° 23.578° 156.422°
10
4
0
1
t1
90°
2 180°
t2
In general, therefore, keep in mind that Equations (13.15) and
(13.16) will provide an angle with a magnitude between 0° and 90°.
b. Eq. (13.10): a qt, and so t a/q. However, a must be in radians.
Thus,
p
a (rad) (23.578°) 0.411 rad
180°
FIG. 13.19
Example 13.9.
and
0.411 rad
a
t1 1.09 ms
377 rad/s
q
For the second intersection,
p
a (rad) (156.422°) 2.73 rad
180°
2.73 rad
a
t2 7.24 ms
377 rad/s
q
FIG. 13.20
Example 13.10, horizontal axis in degrees.
The sine wave can also be plotted against time on the horizontal
axis. The time period for each interval can be determined from t a/q,
but the most direct route is simply to find the period T from T 1/f and
break it up into the required intervals. This latter technique will be
demonstrated in Example 13.10.
Before reviewing the example, take special note of the relative simplicity of the mathematical equation that can represent a sinusoidal
waveform. Any alternating waveform whose characteristics differ from
those of the sine wave cannot be represented by a single term, but may
require two, four, six, or perhaps an infinite number of terms to be represented accurately. Additional description of nonsinusoidal waveforms
can be found in Chapter 24.
EXAMPLE 13.10 Sketch e 10 sin 314t with the abscissa
a. angle (a) in degrees.
b. angle (a) in radians.
c. time (t) in seconds.
FIG. 13.21
Example 13.10, horizontal axis in radians.
Solutions:
a. See Fig 13.20. (Note that no calculations are required.)
b. See Fig. 13.21. (Once the relationship between degrees and radians
is understood, there is again no need for calculations.)
PHASE RELATIONS

523
2p
2p
c. 360°: T 20 ms
314
q
T
20 ms
180°: 10 ms
2
2
T
20 ms
90°: 5 ms
4
4
T
20 ms
30°: 1.67 ms
12
12
FIG. 13.22
Example 13.10, horizontal axis in
milliseconds.
See Fig. 13.22.
EXAMPLE 13.11 Given i 6 103 sin 1000t, determine i at t 2 ms.
Solution:
a qt 1000t (1000 rad/s)(2 103 s) 2 rad
180°
a (°) (2 rad) 114.59°
p rad
i (6 103)(sin 114.59°)
(6 mA)(0.9093) 5.46 mA
Am
( – )
Am sin 13.5
(2 – )
PHASE RELATIONS
Thus far, we have considered only sine waves that have maxima at p/2
and 3p/2, with a zero value at 0, p, and 2p, as shown in Fig. 13.21. If the
waveform is shifted to the right or left of 0°, the expression becomes
Am sin(qt v)
(13.17)
where v is the angle in degrees or radians that the waveform has been
shifted.
If the waveform passes through the horizontal axis with a positivegoing (increasing with time) slope before 0°, as shown in Fig. 13.23,
the expression is
Am sin(qt v)
(13.18)
At qt a 0°, the magnitude is determined by Am sin v. If the waveform passes through the horizontal axis with a positive-going slope
after 0°, as shown in Fig. 13.24, the expression is
Am sin(qt v)
Am
(p + v)
v
(2p + v)
– Am sin v
FIG. 13.24
Defining the phase shift for a sinusoidal
function that crosses the horizontal axis with
a positive slope after 0°.
(13.19)
And at qt a 0°, the magnitude is Am sin(v), which, by a trigonometric identity, is Am sin v.
If the waveform crosses the horizontal axis with a positive-going slope
90° (p/2) sooner, as shown in Fig. 13.25, it is called a cosine wave; that is,
FIG. 13.23
Defining the phase shift for a sinusoidal
function that crosses the horizontal axis with
a positive slope before 0°.
p
sin(qt 90°) sin qt cos qt
2
(13.20)
cos Am
p
2
–p
2
90°
0
sin p
3p
2
2p
FIG. 13.25
Phase relationship between a sine wave and a
cosine wave.
524

SINUSOIDAL ALTERNATING WAVEFORMS
p
sin qt cos(qt 90°) cos qt 2
or
+ cos α
– sin α
+ sin α
– cos α
(13.21)
The terms lead and lag are used to indicate the relationship
between two sinusoidal waveforms of the same frequency plotted on
the same set of axes. In Fig. 13.25, the cosine curve is said to lead
the sine curve by 90°, and the sine curve is said to lag the cosine
curve by 90°. The 90° is referred to as the phase angle between the
two waveforms. In language commonly applied, the waveforms are
out of phase by 90°. Note that the phase angle between the two
waveforms is measured between those two points on the horizontal
axis through which each passes with the same slope. If both waveforms cross the axis at the same point with the same slope, they are
in phase.
The geometric relationship between various forms of the sine and
cosine functions can be derived from Fig. 13.26. For instance, starting
at the sin a position, we find that cos a is an additional 90° in the counterclockwise direction. Therefore, cos a sin(a 90°). For sin a
we must travel 180° in the counterclockwise (or clockwise) direction so
that sin a sin(a 180°), and so on, as listed below:
FIG. 13.26
Graphic tool for finding the relationship
between specific sine and cosine functions.
cos a sin(a 90°)
sin a cos(a 90°)
sin a sin(a 180°)
cos a sin(a 270°) sin(a 90°)
etc.
(13.22)
In addition, one should be aware that
sin(a) sin a
cos(a) cos a
(13.23)
If a sinusoidal expression should appear as
e Em sin qt
the negative sign is associated with the sine portion of the expression,
not the peak value Em. In other words, the expression, if not for convenience, would be written
e Em(sin qt)
Since
sin qt sin(qt 180°)
the expression can also be written
e Em sin(qt 180°)
revealing that a negative sign can be replaced by a 180° change in
phase angle ( or ); that is,
e Em sin qt Em sin(qt 180°)
Em sin(qt 180°)
A plot of each will clearly show their equivalence. There are, therefore, two correct mathematical representations for the functions.
PHASE RELATIONS
The phase relationship between two waveforms indicates which
one leads or lags, and by how many degrees or radians.
EXAMPLE 13.12 What is the phase relationship between the sinusoidal waveforms of each of the following sets?
a. v 10 sin(qt 30°)
i 5 sin(qt 70°)
b. i 15 sin(qt 60°)
v 10 sin(qt 20°)
c. i 2 cos(qt 10°)
v 3 sin(qt 10°)
d. i sin(qt 30°)
v 2 sin(qt 10°)
e. i 2 cos(qt 60°)
v 3 sin(qt 150°)
Solutions:
a. See Fig. 13.27.
i leads v by 40°, or v lags i by 40°.
v
10
5
40° 30°
0
3
2
2
2
t
70°
i
FIG. 13.27
Example 13.12; i leads v by 40°.
b. See Fig. 13.28.
i leads v by 80°, or v lags i by 80°.
i
v
10 15
–
2
0
60°
2
3
2
20°
80°
FIG. 13.28
Example 13.12; i leads v by 80°.
2
t

525
526

SINUSOIDAL ALTERNATING WAVEFORMS
c. See Fig. 13.29.
i 2 cos(qt 10°) 2 sin(qt 10° 90°)
2 sin(qt 100°)
i leads v by 110°, or v lags i by 110°.
i
2
–
3
v
100°
0
2
10° 2
3
2
2
t
110°
FIG. 13.29
Example 13.12; i leads v by 110°.
d. See Fig. 13.30.
Note
sin(qt 30°) sin(qt 30° 180°)
sin(qt 150°)
v leads i by 160°, or i lags v by 160°.
v
i
2
–
2
0
2
10°
5
2
1
3
2
150°
160°
3
2
t
200°
360°
FIG. 13.30
Example 13.12; v leads i by 160°.
Or using
Note
sin(qt 30°) sin(qt 30° 180°)
sin(qt 210°)
i leads v by 200°, or v lags i by 200°.
e. See Fig. 13.31.
By choice
i 2 cos(qt 60°) 2 cos(qt 60° 180°)
2 cos(qt 240°)
v
2
–
2
0
2
150°
3
i
3
2
2
5
2
FIG. 13.31
Example 13.12; v and i are in phase.
3
t
AVERAGE VALUE

527
cos a sin(a 90°)
However,
2 cos(qt 240°) 2 sin(qt 240° 90°)
2 sin(qt 150°)
v and i are in phase.
so that
Phase Measurements
The hookup procedure for using an oscilloscope to measure phase
angles is covered in detail in Section 15.13. However, the equation for
determining the phase angle can be introduced using Fig. 13.32. First,
note that each sinusoidal function has the same frequency, permitting
the use of either waveform to determine the period. For the waveform
chosen in Fig. 13.32, the period encompasses 5 divisions at 0.2 ms/div.
The phase shift between the waveforms (irrespective of which is leading or lagging) is 2 divisions. Since the full period represents a cycle of
360°, the following ratio [from which Equation (13.24) can be derived]
can be formed:
phase shift (no. of div.)
v 360°
T (no. of div.)
i
θ
T
360°
v
T (no. of div.)
phase shift (no. of div.)
and
e
Vertical sensitivity = 2 V/div.
Horizontal sensitivity = 0.2 ms/div.
(13.24)
FIG. 13.32
Finding the phase angle between waveforms
using a dual-trace oscilloscope.
Substituting into Eq. (13.24) will result in
(2 div.)
v 360° 144°
(5 div.)
and e leads i by 144°.
13.6
AVERAGE VALUE
Even though the concept of the average value is an important one in
most technical fields, its true meaning is often misunderstood. In Fig.
13.33(a), for example, the average height of the sand may be required
to determine the volume of sand available. The average height of the
sand is that height obtained if the distance from one end to the other
is maintained while the sand is leveled off, as shown in Fig. 13.33(b).
The area under the mound of Fig. 13.33(a) will then equal the area
under the rectangular shape of Fig. 13.33(b) as determined by A b h. Of course, the depth (into the page) of the sand must be the
same for Fig. 13.33(a) and (b) for the preceding conclusions to have
any meaning.
In Fig. 13.33 the distance was measured from one end to the other.
In Fig. 13.34(a) the distance extends beyond the end of the original pile
of Fig. 13.33. The situation could be one where a landscaper would like
to know the average height of the sand if spread out over a distance
such as defined in Fig. 13.34(a). The result of an increased distance is
as shown in Fig. 13.34(b). The average height has decreased compared
to Fig. 13.33. Quite obviously, therefore, the longer the distance, the
lower is the average value.
Height
Sand
Distance
(a)
Height
Average height
Sand
Same
distance
(b)
FIG. 13.33
Defining average value.
528

SINUSOIDAL ALTERNATING WAVEFORMS
If the distance parameter includes a depression, as shown in Fig.
13.35(a), some of the sand will be used to fill the depression, resulting
in an even lower average value for the landscaper, as shown in Fig.
13.35(b). For a sinusoidal waveform, the depression would have the
same shape as the mound of sand (over one full cycle), resulting in an
average value at ground level (or zero volts for a sinusoidal voltage over
one full period).
After traveling a considerable distance by car, some drivers like to
calculate their average speed for the entire trip. This is usually done by
dividing the miles traveled by the hours required to drive that distance.
For example, if a person traveled 225 mi in 5 h, the average speed was
225 mi/5 h, or 45 mi/h. This same distance may have been traveled at
various speeds for various intervals of time, as shown in Fig. 13.36.
By finding the total area under the curve for the 5 h and then dividing the area by 5 h (the total time for the trip), we obtain the same result
of 45 mi/h; that is,
Height
Sand
Distance
(a)
Height
Average height
Sand
Same
distance
(b)
area under curve
Average speed length of curve
FIG. 13.34
Effect of distance (length) on average value.
(13.25)
A1 A2
Average speed 5h
(60 mi/h)(2 h) (50 mi/h)(2.5 h)
5h
225
mi/h
5
Height
Sand
Ground level
Distance
45 mi/h
(a)
Equation (13.25) can be extended to include any variable quantity, such
as current or voltage, if we let G denote the average value, as follows:
Height
algebraic sum of areas
G (average value) length of curve
Average height
Sand
Same
distance
(b)
FIG. 13.35
Effect of depressions (negative excursions) on
average value.
(13.26)
Speed (mi/h)
70
60
50
40
30
20
10
0
Average speed
A1
1
A2
2
3
Lunch break
4
5
6
t (h)
FIG. 13.36
Plotting speed versus time for an automobile excursion.
The algebraic sum of the areas must be determined, since some area
contributions will be from below the horizontal axis. Areas above the
axis will be assigned a positive sign, and those below, a negative sign.
A positive average value will then be above the axis, and a negative
value, below.
The average value of any current or voltage is the value indicated on
a dc meter. In other words, over a complete cycle, the average value is
AVERAGE VALUE

529
the equivalent dc value. In the analysis of electronic circuits to be considered in a later course, both dc and ac sources of voltage will be
applied to the same network. It will then be necessary to know or determine the dc (or average value) and ac components of the voltage or current in various parts of the system.
EXAMPLE 13.13 Determine the average value of the waveforms of
Fig. 13.37.
v2
v1
(Square wave)
14 V
10 V
0
1
2
3
4
t (ms)
0
1
2
3
4
t (ms)
–6 V
–10 V
(b)
(a)
FIG. 13.37
Example 13.13.
Solutions:
a. By inspection, the area above the axis equals the area below over
one cycle, resulting in an average value of zero volts. Using Eq.
(13.26):
(10 V)(1 ms) (10 V)(1 ms)
G 2 ms
0
0V
2 ms
b. Using Eq. (13.26):
(14 V)(1 ms) (6 V)(1 ms)
G 2 ms
8V
14 V 6 V
4 V
2
2
14 V
4V
0
v2 v1 4 V
EXAMPLE 13.14 Find the average values of the following waveforms
over one full cycle:
a. Fig. 13.39.
b. Fig. 13.40.
1
2
3
4
t (ms)
–6 V
as shown in Fig. 13.38.
In reality, the waveform of Fig. 13.37(b) is simply the square wave
of Fig. 13.37(a) with a dc shift of 4 V; that is,
FIG. 13.38
Defining the average value for the waveform
of Fig. 13.37(b).
v (V)
1 cycle
3
0
4
8
t (ms)
–1
FIG. 13.39
Example 13.14, part (a).
530

SINUSOIDAL ALTERNATING WAVEFORMS
i (A)
1 cycle
4
6
0
2
–2
8
4
10
t (ms)
–10
FIG. 13.40
Example 13.14, part (b).
vav (V)
0 1V
1
8 t (ms)
dc voltmeter (between 0 and 8 ms)
FIG. 13.41
The response of a dc meter to the waveform of
Fig. 13.39.
iav (A)
0
–1.6
10
t (ms)
– –1.6 +
dc ammeter (between 0 and 10 ms)
FIG. 13.42
The response of a dc meter to the waveform of
Fig. 13.40.
Solutions:
(3 V)(4 ms) (1 V)(4 ms)
12 V 4 V
a. G 1 V
8 ms
8
Note Fig. 13.41.
(10 V)(2 ms) (4 V)(2 ms) (2 V)(2 ms)
b. G 10 ms
16 V
20 V 8 V 4 V
1.6 V
10
10
Note Fig. 13.42.
We found the areas under the curves in the preceding example by
using a simple geometric formula. If we should encounter a sine wave
or any other unusual shape, however, we must find the area by some
other means. We can obtain a good approximation of the area by
attempting to reproduce the original wave shape using a number of
small rectangles or other familiar shapes, the area of which we already
know through simple geometric formulas. For example,
the area of the positive (or negative) pulse of a sine wave is 2Am.
Approximating this waveform by two triangles (Fig. 13.43), we obtain
(using area 1/2 base height for the area of a triangle) a rough idea
of the actual area:
b
h






FIG. 13.43
Approximating the shape of the positive pulse
of a sinusoidal waveform with two right
triangles.
1
1 p
p
Area shaded 2 bh 2 (Am) Am
2
2 2
2
1.58Am
A closer approximation might be a rectangle with two similar triangles (Fig. 13.44):
2
1
p
p
p
Area Am 2 bh Am Am pAm
2
3
3
3
3
2.094Am
FIG. 13.44
A better approximation for the shape of the
positive pulse of a sinusoidal waveform.
which is certainly close to the actual area. If an infinite number of
forms were used, an exact answer of 2Am could be obtained. For irregular waveforms, this method can be especially useful if data such as the
average value are desired.
The procedure of calculus that gives the exact solution 2Am is
known as integration. Integration is presented here only to make the
AVERAGE VALUE

531
method recognizable to the reader; it is not necessary to be proficient in
its use to continue with this text. It is a useful mathematical tool, however, and should be learned. Finding the area under the positive pulse of
a sine wave using integration, we have
Area p
Am sin a da
0
where ∫ is the sign of integration, 0 and p are the limits of integration,
Am sin a is the function to be integrated, and da indicates that we are
integrating with respect to a.
Integrating, we obtain
Area Am[cos a]p0
Am(cos p cos 0°)
Am[1 (1)] Am(2)
Am
Area 2Am
(13.27)
p
0
Since we know the area under the positive (or negative) pulse, we
can easily determine the average value of the positive (or negative)
region of a sine wave pulse by applying Eq. (13.26):
2Am
G
p
and
G
G 0.637Am
0
Am
(13.28)
p
For the waveform of Fig. 13.45,
2Am
(2Am /2)
G
p
p/2
FIG. 13.45
Finding the average value of one-half the
positive pulse of a sinusoidal waveform.
(average the same
as for a full pulse)
1 cycle
EXAMPLE 13.15 Determine the average value of the sinusoidal
waveform of Fig. 13.46.
Am
π
0
Solution:
By inspection it is fairly obvious that
the average value of a pure sinusoidal waveform over one full cycle is
zero.
Am
2π
α
FIG. 13.46
Example 13.15.
Eq. (13.26):
2Am 2Am
G 0 V
2p
v
+2 mV
EXAMPLE 13.16 Determine the average value of the waveform of
Fig. 13.47.
Solution: The peak-to-peak value of the sinusoidal function is
16 mV 2 mV 18 mV. The peak amplitude of the sinusoidal waveform is, therefore, 18 mV/2 9 mV. Counting down 9 mV from 2 mV
(or 9 mV up from 16 mV) results in an average or dc level of 7 mV,
as noted by the dashed line of Fig. 13.47.
t
0
– 16 mV
FIG. 13.47
Example 13.16.

532
SINUSOIDAL ALTERNATING WAVEFORMS
v (V)
EXAMPLE 13.17 Determine the average value of the waveform of
Fig. 13.48.
1 cycle
10
Sine wave
p
0
Solution:
2p
1
FIG. 13.48
Example 13.17.
2(10 V)
2Am 0
G 3.18 V
2p
2p
EXAMPLE 13.18 For the waveform of Fig. 13.49, determine whether
the average value is positive or negative, and determine its approximate
value.
Solution: From the appearance of the waveform, the average value
is positive and in the vicinity of 2 mV. Occasionally, judgments of this
type will have to be made.
v (mV)
10 mV
t
0
Instrumentation
The dc level or average value of any waveform can be found using a
digital multimeter (DMM) or an oscilloscope. For purely dc circuits,
simply set the DMM on dc, and read the voltage or current levels.
Oscilloscopes are limited to voltage levels using the sequence of steps
listed below:
FIG. 13.49
Example 13.18.
1. First choose GND from the DC-GND-AC option list associated
with each vertical channel. The GND option blocks any signal to
which the oscilloscope probe may be connected from entering the
oscilloscope and responds with just a horizontal line. Set the
resulting line in the middle of the vertical axis on the horizontal
axis, as shown in Fig. 13.50(a).
Shift = 2.5 div.
(a)
Vertical sensitivity = 50 mV/div.
(b)
FIG. 13.50
Using the oscilloscope to measure dc voltages: (a) setting the GND condition;
(b) the vertical shift resulting from a dc voltage when shifted to the DC option.
2. Apply the oscilloscope probe to the voltage to be measured (if
not already connected), and switch to the DC option. If a dc voltage is present, the horizontal line will shift up or down, as
demonstrated in Fig. 13.50(b). Multiplying the shift by the vertical sensitivity will result in the dc voltage. An upward shift is a
positive voltage (higher potential at the red or positive lead of the
oscilloscope), while a downward shift is a negative voltage
(lower potential at the red or positive lead of the oscilloscope).
AVERAGE VALUE
In general,
Vdc (vertical shift in div.) (vertical sensitivity in V/div.)
(13.29)
For the waveform of Fig. 13.50(b),
Vdc (2.5 div.)(50 mV/div.) 125 mV
The oscilloscope can also be used to measure the dc or average level
of any waveform using the following sequence:
1. Using the GND option, reset the horizontal line to the middle of
the screen.
2. Switch to AC (all dc components of the signal to which the probe
is connected will be blocked from entering the oscilloscope—
only the alternating, or changing, components will be displayed).
Note the location of some definitive point on the waveform, such
as the bottom of the half-wave rectified waveform of Fig.
13.51(a); that is, note its position on the vertical scale. For the
future, whenever you use the AC option, keep in mind that the
computer will distribute the waveform above and below the horizontal axis such that the average value is zero; that is, the area
above the axis will equal the area below.
3. Then switch to DC (to permit both the dc and the ac components
of the waveform to enter the oscilloscope), and note the shift in
the chosen level of part 2, as shown in Fig. 13.51(b). Equation
(13.29) can then be used to determine the dc or average value of
the waveform. For the waveform of Fig. 13.51(b), the average
value is about
Vav Vdc (0.9 div.)(5 V/div.) 4.5 V
Reference
level
(a)
Shift = 0.9 div.
(b)
FIG. 13.51
Determining the average value of a nonsinusoidal waveform using the
oscilloscope: (a) vertical channel on the ac mode; (b) vertical channel on the
dc mode.
The procedure outlined above can be applied to any alternating
waveform such as the one in Fig. 13.49. In some cases the average
value may require moving the starting position of the waveform under
the AC option to a different region of the screen or choosing a higher
voltage scale. DMMs can read the average or dc level of any waveform
by simply choosing the appropriate scale.

533
534

SINUSOIDAL ALTERNATING WAVEFORMS
13.7 EFFECTIVE VALUES
This section will begin to relate dc and ac quantities with respect to
the power delivered to a load. It will help us determine the amplitude
of a sinusoidal ac current required to deliver the same power as a
particular dc current. The question frequently arises, How is it possible for a sinusoidal ac quantity to deliver a net power if, over a full
cycle, the net current in any one direction is zero (average value 0)? It would almost appear that the power delivered during the positive portion of the sinusoidal waveform is withdrawn during the negative portion, and since the two are equal in magnitude, the net
power delivered is zero. However, understand that irrespective of
direction, current of any magnitude through a resistor will deliver
power to that resistor. In other words, during the positive or negative
portions of a sinusoidal ac current, power is being delivered at each
instant of time to the resistor. The power delivered at each instant
will, of course, vary with the magnitude of the sinusoidal ac current,
but there will be a net flow during either the positive or the negative
pulses with a net flow over the full cycle. The net power flow will
equal twice that delivered by either the positive or the negative
regions of sinusoidal quantity.
A fixed relationship between ac and dc voltages and currents can be
derived from the experimental setup shown in Fig. 13.52. A resistor in
a water bath is connected by switches to a dc and an ac supply. If switch
1 is closed, a dc current I, determined by the resistance R and battery
voltage E, will be established through the resistor R. The temperature
reached by the water is determined by the dc power dissipated in the
form of heat by the resistor.
iac
Switch 1
Switch 2
R
e
ac generator
E
Idc
dc source
FIG. 13.52
An experimental setup to establish a relationship between dc and ac quantities.
If switch 2 is closed and switch 1 left open, the ac current through
the resistor will have a peak value of Im. The temperature reached by
the water is now determined by the ac power dissipated in the form of
heat by the resistor. The ac input is varied until the temperature is the
same as that reached with the dc input. When this is accomplished, the
average electrical power delivered to the resistor R by the ac source is
the same as that delivered by the dc source.
The power delivered by the ac supply at any instant of time is
Pac (iac)2R (Im sin qt)2R (I 2m sin2qt)R
but
1
sin2 qt (1 cos 2qt)
2
(trigonometric identity)
EFFECTIVE VALUES
Therefore,
1
Pac I m2 (1 cos 2qt) R
2
and
I m2 R
I m2 R
Pac cos 2qt
2
2
(13.30)
The average power delivered by the ac source is just the first term,
since the average value of a cosine wave is zero even though the wave
may have twice the frequency of the original input current waveform.
Equating the average power delivered by the ac generator to that delivered by the dc source,
Pav(ac) Pdc
I m2 R
I 2dc R and
2
or
Im 2Idc
Im
Idc 0.707Im
2
which, in words, states that
the equivalent dc value of a sinusoidal current or voltage is 1/ 2 or
0.707 of its maximum value.
The equivalent dc value is called the effective value of the sinusoidal
quantity.
In summary,
or
and
or
Ieq(dc) Ieff 0.707Im
(13.31)
Im 2Ieff 1.414Ieff
(13.32)
Eeff 0.707Em
(13.33)
Em 2Eeff 1.414Eeff
(13.34)
As a simple numerical example, it would require an ac current with
a peak value of 2(10) 14.14 A to deliver the same power to the
resistor in Fig. 13.52 as a dc current of 10 A. The effective value of any
quantity plotted as a function of time can be found by using the following equation derived from the experiment just described:
i (t) dt
T
or
2
Ieff T
(13.35)
Ieff area (i2(t))
T
(13.36)
0

535
536

SINUSOIDAL ALTERNATING WAVEFORMS
which, in words, states that to find the effective value, the function i(t)
must first be squared. After i(t) is squared, the area under the curve is
found by integration. It is then divided by T, the length of the cycle or
the period of the waveform, to obtain the average or mean value of the
squared waveform. The final step is to take the square root of the mean
value. This procedure gives us another designation for the effective
value, the root-mean-square (rms) value.
EXAMPLE 13.19 Find the effective values of the sinusoidal waveform in each part of Fig. 13.53.
i (mA)
v
i (mA)
12
12
0
1s
t
169.7 V
1s
0
(a)
2s
t
t
(b)
(c)
FIG. 13.53
Example 13.19.
Solution: For part (a), Ieff 0.707(12 103 A) 8.484 mA.
For part (b), again Ieff 8.484 mA. Note that frequency did not
change the effective value in (b) above as compared to (a). For part
(c), Veff 0.707(169.73 V) 120 V, the same as available from a
home outlet.
EXAMPLE 13.20 The 120-V dc source of Fig. 13.54(a) delivers
3.6 W to the load. Determine the peak value of the applied voltage
(Em) and the current (Im) if the ac source [Fig. 13.54(b)] is to
deliver the same power to the load.
Im
Idc
iac
e
E
120 V
Em
P = 3.6 W
Load
+
P = 3.6 W
Load
–
(a)
(b)
FIG. 13.54
Example 13.20.
EFFECTIVE VALUES
Solution:
537
v (V)
1 cycle
Pdc VdcIdc
3
Pdc
3.6 W
Idc 30 mA
120 V
Vdc
and

0
Im 2Idc (1.414)(30 mA) 42.42 mA
Em 2Edc (1.414)(120 V) 169.68 V
4
8
t (s)
–1
FIG. 13.55
Example 13.21.
EXAMPLE 13.21 Find the effective or rms value of the waveform of
Fig. 13.55.
v2 (V)
Solution:
v2 (Fig. 13.56):
(9)(4) (1)(4)
8
Veff 9
40
2.236 V
8
EXAMPLE 13.22 Calculate the effective value of the voltage of Fig.
13.57.
(– 1)2 = 1
1
0
v (V)
1 cycle
4
6
8
10
t (s)
–10
FIG. 13.57
Example 13.22.
Solution:
v2 (Fig. 13.58):
Veff (100)(2) (16)(2) (4)(2)
10
240
10
4.899 V
v2 (V)
100
16
4
0
2
4
8
FIG. 13.56
The squared waveform of Fig. 13.55.
4
0
–2
4
6
8
FIG. 13.58
The squared waveform of Fig. 13.57.
10
t (s)
t (s)
538

SINUSOIDAL ALTERNATING WAVEFORMS
v (V)
EXAMPLE 13.23 Determine the average and effective values of the
square wave of Fig. 13.59.
40
0
– 40
10
20
t (ms)
1 cycle
Solution: By inspection, the average value is zero.
v2 (Fig. 13.60):
(1600)(10 103) (1600)(10 103)
Veff 20 103
FIG. 13.59
Example 13.23.
32,000 103
1600
20 103
Veff 40 V
v2 (V)
(the maximum value of the waveform of Fig. 13.60)
1600
0
10
20
t (ms)
FIG. 13.60
The squared waveform of Fig. 13.59.
The waveforms appearing in these examples are the same as those
used in the examples on the average value. It might prove interesting to
compare the effective and average values of these waveforms.
The effective values of sinusoidal quantities such as voltage or current will be represented by E and I. These symbols are the same as
those used for dc voltages and currents. To avoid confusion, the peak
value of a waveform will always have a subscript m associated with it:
Im sin qt. Caution: When finding the effective value of the positive
pulse of a sine wave, note that the squared area is not simply (2Am)2 4A2m; it must be found by a completely new integration. This will
always be the case for any waveform that is not rectangular.
A unique situation arises if a waveform has both a dc and an ac component that may be due to a source such as the one in Fig. 13.61. The
combination appears frequently in the analysis of electronic networks
where both dc and ac levels are present in the same system.
vT
+
+
3 sin ω t
–
6V
vT
7.5 V
6V
4.5 V
–
0
t
FIG. 13.61
Generation and display of a waveform having a dc and an ac component.
The question arises, What is the effective value of the voltage vT?
One might be tempted to simply assume that it is the sum of the effective values of each component of the waveform; that is, VT (eff) 0.7071(1.5 V) 6 V 1.06 V 6 V 7.06 V. However, the rms
value is actually determined by
2
Veff V 2dc V ac
(rms)
which for the above example is
Veff (6 V)2 (1.06 V)2
37.124 V
6.1 V
(13.37)
ac METERS AND INSTRUMENTS
This result is noticeably less than the above solution. The development
of Eq. (13.37) can be found in Chapter 24.
Instrumentation
It is important to note whether the DMM in use is a true rms meter or
simply a meter where the average value is calibrated (as described in
the next section) to indicate the rms level. A true rms meter will read
the effective value of any waveform (such as Figs. 13.49 and 13.61)
and is not limited to only sinusoidal waveforms. Since the label true
rms is normally not placed on the face of the meter, it is prudent to
check the manual if waveforms other than purely sinusoidal are to be
encountered.
13.8
ac METERS AND INSTRUMENTS
The d’Arsonval movement employed in dc meters can also be used to
measure sinusoidal voltages and currents if the bridge rectifier of Fig.
13.62 is placed between the signal to be measured and the average reading movement.
The bridge rectifier, composed of four diodes (electronic switches),
will convert the input signal of zero average value to one having an
average value sensitive to the peak value of the input signal. The conversion process is well described in most basic electronics texts. Fundamentally, conduction is permitted through the diodes in such a manner as to convert the sinusoidal input of Fig. 13.63(a) to one having the
appearance of Fig. 13.63(b). The negative portion of the input has been
effectively “flipped over” by the bridge configuration. The resulting
waveform of Fig. 13.63(b) is called a full-wave rectified waveform.
vi
vmovement
Vm
0
Vm
2 0
Vdc = 0.637Vm
2
–Vm
(a)
(b)
FIG. 13.63
(a) Sinusoidal input; (b) full-wave rectified signal.
The zero average value of Fig. 13.63(a) has been replaced by a pattern having an average value determined by
2Vm
4Vm
2Vm 2Vm
G 0.637Vm
p
2p
2p
The movement of the pointer will therefore be directly related to the
peak value of the signal by the factor 0.637.
Forming the ratio between the rms and dc levels will result in
0.707Vm
Vrms
1.11
0.637Vm
Vdc
+
vmovement
vi
–
+
–
FIG. 13.62
Full-wave bridge rectifier.

539
540

SINUSOIDAL ALTERNATING WAVEFORMS
revealing that the scale indication is 1.11 times the dc level measured
by the movement; that is,
Meter indication 1.11 (dc or average value)
Some ac meters use a half-wave rectifier arrangement that results in
the waveform of Fig. 13.64, which has half the average value of Fig.
13.63(b) over one full cycle. The result is
vmovement
Vm
Vdc = 0.318Vm
full-wave (13.38)
2
FIG. 13.64
Half-wave rectified signal.
FIG. 13.65
Electrodynamometer movement. (Courtesy of
Weston Instruments, Inc.)
Meter indication 2.22 (dc or average value)
half-wave (13.39)
A second movement, called the electrodynamometer movement
(Fig. 13.65), can measure both ac and dc quantities without a change in
internal circuitry. The movement can, in fact, read the effective value of
any periodic or nonperiodic waveform because a reversal in current
direction reverses the fields of both the stationary and the movable
coils, so the deflection of the pointer is always up-scale.
The VOM, introduced in Chapter 2, can be used to measure both dc
and ac voltages using a d’Arsonval movement and the proper switching
networks. That is, when the meter is used for dc measurements, the dial
setting will establish the proper series resistance for the chosen scale
and will permit the appropriate dc level to pass directly to the movement. For ac measurements, the dial setting will introduce a network
that employs a full- or half-wave rectifier to establish a dc level. As discussed above, each setting is properly calibrated to indicate the desired
quantity on the face of the instrument.
EXAMPLE 13.24 Determine the reading of each meter for each situation of Fig. 13.66(a) and (b).
d’Arsonval
movement
rms scale
(full-wave
rectifier)
dc
ac
+
+
20 V
–
Vm = 20 V
–
Voltmeter
(1)
(2)
(a)
Electrodynamometer
movement
rms scale
dc
+
+
25 V
e = 15 sin 200t
–
–
Voltmeter
(1)
(b)
FIG. 13.66
Example 13.24.
(2)
ac METERS AND INSTRUMENTS
Solution:

541
For Fig. 13.66(a), situation (1): By Eq. (13.38),
Meter indication 1.11(20 V) 22.2 V
For Fig. 13.66(a), situation (2):
Vrms 0.707Vm 0.707(20 V) 14.14 V
For Fig. 13.66(b), situation (1):
Vrms Vdc 25 V
For Fig. 13.66(b), situation (2):
Vrms 0.707Vm 0.707(15 V) 10.6 V
Most DMMs employ a full-wave rectification system to convert the
input ac signal to one with an average value. In fact, for the DMM of
Fig. 2.27, the same scale factor of Eq. (13.38) is employed; that is, the
average value is scaled up by a factor of 1.11 to obtain the rms value.
In digital meters, however, there are no moving parts such as in the
d’Arsonval or electrodynamometer movements to display the signal
level. Rather, the average value is sensed by a multiprocessor integrated
circuit (IC), which in turn determines which digits should appear on the
digital display.
Digital meters can also be used to measure nonsinusoidal signals,
but the scale factor of each input waveform must first be known (normally provided by the manufacturer in the operator’s manual). For
instance, the scale factor for an average responding DMM on the ac rms
scale will produce an indication for a square-wave input that is 1.11
times the peak value. For a triangular input, the response is 0.555 times
the peak value. Obviously, for a sine wave input, the response is 0.707
times the peak value.
For any instrument, it is always good practice to read (if only briefly)
the operator’s manual if it appears that you will use the instrument on a
regular basis.
For frequency measurements, the frequency counter of Fig. 13.67
provides a digital readout of sine, square, and triangular waves from
5 Hz to 100 MHz at input levels from 30 mV to 42 V. Note the relative
simplicity of the panel and the high degree of accuracy available.
The Amp-Clamp® of Fig. 13.68 is an instrument that can measure
alternating current in the ampere range without having to open the circuit. The loop is opened by squeezing the “trigger”; then it is placed
around the current-carrying conductor. Through transformer action, the
level of current in rms units will appear on the appropriate scale. The
accuracy of this instrument is 3% of full scale at 60 Hz, and its scales
have maximum values ranging from 6 A to 300 A. The addition of two
leads, as indicated in the figure, permits its use as both a voltmeter and
an ohmmeter.
One of the most versatile and important instruments in the electronics industry is the oscilloscope, which has already been introduced in
this chapter. It provides a display of the waveform on a cathode-ray
tube to permit the detection of irregularities and the determination of
quantities such as magnitude, frequency, period, dc component, and so
on. The analog oscilloscope of Fig. 13.69 can display two waveforms at
the same time (dual-channel) using an innovative interface (front
panel). It employs menu buttons to set the vertical and horizontal scales
by choosing from selections appearing on the screen. One can also store
up to four measurement setups for future use.
FIG. 13.67
Frequency counter. (Courtesy of Tektronix,
Inc.)
FIG. 13.68
Amp-Clamp®. (Courtesy of Simpson
Instruments, Inc.)
FIG. 13.69
Dual-channel oscilloscope. (Courtesy of
Tektronix, Inc.)
542

SINUSOIDAL ALTERNATING WAVEFORMS
A student accustomed to watching TV might be confused when
first introduced to an oscilloscope. There is, at least initially, an
assumption that the oscilloscope is generating the waveform on the
screen—much like a TV broadcast. However, it is important to
clearly understand that
an oscilloscope displays only those signals generated elsewhere and
connected to the input terminals of the oscilloscope. The absence of
an external signal will simply result in a horizontal line on the screen
of the scope.
On most modern-day oscilloscopes, there is a switch or knob with
the choice DC/GND/AC, as shown in Fig. 13.70(a), that is often
ignored or treated too lightly in the early stages of scope utilization.
The effect of each position is fundamentally as shown in Fig. 13.70(b).
In the DC mode the dc and ac components of the input signal can pass
directly to the display. In the AC position the dc input is blocked by the
capacitor, but the ac portion of the signal can pass through to the
screen. In the GND position the input signal is prevented from reaching
the scope display by a direct ground connection, which reduces the
scope display to a single horizontal line.
AC
Oscilloscope
display
AC
GND
DC
GND
DC
(a)
Input signal
(b)
FIG. 13.70
AC-GND-DC switch for the vertical channel of an oscilloscope.
13.9
APPLICATIONS
(120 V at 60 Hz) versus (220 V at 50 Hz)
In North and South America the most common available ac supply is
120 V at 60 Hz, while in Europe and the Eastern countries it is 220 V
at 50 Hz. The choices of rms value and frequency were obviously made
carefully because they have such an important impact on the design and
operation of so many systems.
The fact that the frequency difference is only 10 Hz reveals that
there was agreement on the general frequency range that should be used
for power generation and distribution. History suggests that the question of frequency selection was originally focused on that frequency
that would not exhibit flicker in the incandescent lamps available in
those days. Technically, however, there really wouldn’t be a noticeable
difference between 50 and 60 cycles per second based on this criterion.
Another important factor in the early design stages was the effect of frequency on the size of transformers, which play a major role in power
generation and distribution. Working through the fundamental equations for transformer design, you will find that the size of a transformer
is inversely proportional to frequency. The result is that transformers
operating at 50 Hz must be larger (on a purely mathematical basis about
17% larger) than those operating at 60 Hz. You will therefore find that
transformers designed for the international market where they can oper-
APPLICATIONS
ate on 50 Hz or 60 Hz are designed around the 50-Hz frequency. On the
other side of the coin, however, higher frequencies result in increased
concerns about arcing, increased losses in the transformer core due to
eddy current and hysteresis losses (Chapter 19), and skin effect phenomena (Chapter 19). Somewhere in the discussion we must consider
the fact that 60 Hz is an exact multiple of 60 seconds in a minute and
60 minutes in an hour. Since accurate timing is such a critical part of
our technological design, was this a significant motive in the final
choice? There is also the question about whether the 50 Hz is a result
of the close affinity of this value to the metric system. Keep in mind
that powers of 10 are all powerful in the metric system, with 100 cm in
a meter, 100°C the boiling point of water, and so on. Note that 50 Hz is
exactly half of this special number. All in all, it would seem that both
sides have an argument that would be worth defending. However, in the
final analysis, we must also wonder whether the difference is simply
political in nature.
The difference in voltage between North America and Europe is a
different matter entirely in the sense that the difference is close to
100%. Again, however, there are valid arguments for both sides. There
is no question that larger voltages such as 220 V raise safety issues
beyond those raised by voltages of 120 V. However, when higher voltages are supplied, there is less current in the wire for the same power
demand, permitting the use of smaller conductors—a real money saver.
In addition, motors, compressors, and so on, found in common home
appliances and throughout the industrial community can be smaller in
size. Higher voltages, however, also bring back the concern about arcing effects, insulation requirements, and, due to real safety concerns,
higher installation costs. In general, however, international travelers are
prepared for most situations if they have a transformer that can convert
from their home level to that of the country they plan to visit. Most
equipment (not clocks, of course) can run quite well on 50 Hz or 60 Hz
for most travel periods. For any unit not operating at its design frequency, it will simply have to “work a little harder” to perform the
given task. The major problem for the traveler is not the transformer
itself but the wide variety of plugs used from one country to another.
Each country has its own design for the “female” plug in the wall. For
the three-week tour, this could mean as many as 6 to 10 different plugs
of the type shown in Fig. 13.71. For a 120-V, 60-Hz supply, the plug is
quite standard in appearance with its two spade leads (and possible
ground connection).
In any event, both the 120 V at 60 Hz and the 220 V at 50 Hz are
obviously meeting the needs of the consumer. It is a debate that could
go on at length without an ultimate victor.
Safety Concerns (High Voltages and dc versus ac)
Be aware that any “live” network should be treated with a calculated
level of respect. Electricity in its various forms is not to be feared but
should be employed with some awareness of its potentially dangerous
side effects. It is common knowledge that electricity and water do not
mix (never use extension cords or plug in TVs or radios in the bathroom) because a full 120 V in a layer of water of any height (from a
shallow puddle to a full bath) can be lethal. However, other effects of
dc and ac voltages are less known. In general, as the voltage and current increase, your concern about safety should increase exponentially.

FIG. 13.71
Variety of plugs for a 220-V, 50-Hz
connection.
543
544

SINUSOIDAL ALTERNATING WAVEFORMS
V(volts)
170
120 V rms
ac voltage
20
0
–20
t
tf
FIG 13.72
Interval of time when sinusoidal voltage is
near zero volts.
For instance, under dry conditions, most human beings can survive a
120-V ac shock such as obtained when changing a light bulb, turning
on a switch, and so on. Most electricians have experienced such a jolt
many times in their careers. However, ask an electrician to relate how it
feels to hit 220 V, and the response (if he or she has been unfortunate to
have had such an experience) will be totally different. How often have
you heard of a back-hoe operator hitting a 220-V line and having a fatal
heart attack? Remember, the operator is sitting in a metal container on
a damp ground which provides an excellent path for the resulting current to flow from the line to ground. If only for a short period of time,
with the best environment (rubber-sole shoes, etc.), in a situation where
you can quickly escape the situation, most human beings can also survive a 220-V shock. However, as mentioned above, it is one you will
not quickly forget. For voltages beyond 220 V rms, the chances of survival go down exponentially with increase in voltage. It takes only
about 10 mA of steady current through the heart to put it in defibrillation. In general, therefore, always be sure that the power is disconnected when working on the repair of electrical equipment. Don’t
assume that throwing a wall switch will disconnect the power. Throw
the main circuit breaker and test the lines with a voltmeter before working on the system. Since voltage is a two-point phenomenon, don’t be a
hero and work with one line at at time—accidents happen!
You should also be aware that the reaction to dc voltages is quite different from that to ac voltages. You have probably seen in movies or
comic strips that people are often unable to let go of a hot wire. This is
evidence of the most important difference between the two types of
voltages. As mentioned above, if you happen to touch a “hot” 120-V ac
line, you will probably get a good sting, but you can let go. If it happens to be a “hot” 120-V dc line, you will probably not be able to let
go, and a fatality could occur. Time plays an important role when this
happens, because the longer you are subjected to the dc voltage, the
more the resistance in the body decreases until a fatal current can be
established. The reason that we can let go of an ac line is best demonstrated by carefully examining the 120-V rms, 60-Hz voltage in Fig.
13.72. Since the voltage is oscillating, there is a period of time when the
voltage is near zero or less than, say, 20 V, and is reversing in direction.
Although this time interval is very short, it appears every 8.3 ms and
provides a window to let go.
Now that we are aware of the additional dangers of dc voltages, it is
important to mention that under the wrong conditions, dc voltages as
low as 12 V such as from a car battery can be quite dangerous. If you
happen to be working on a car under wet conditions, or if you are
sweating badly for some reason or, worse yet, wearing a wedding ring
that may have moisture and body salt underneath, touching the positive
terminal may initiate the process whereby the body resistance begins to
drop and serious injury could take place. It is one of the reasons you
seldom see a professional electrician wearing any rings or jewelry—it
is just not worth the risk.
Before leaving this topic of safety concerns, you should also be
aware of the dangers of high-frequency supplies. We are all aware of
what 2.45 GHz at 120 V can do to a meat product in a microwave oven.
As discussed in Chapter 5, it is therefore very important that the seal
around the oven be as tight as possible. However, don’t ever assume
that anything is absolutely perfect in design—so don’t make it a habit
to view the cooking process in the microwave 6 in. from the door on a
APPLICATIONS

545
continuing basis. Find something else to do, and check the food only
when the cooking process is complete. If you ever visit the Empire
State Building, you will notice that you are unable to get close to the
antenna on the dome due to the high-frequency signals being emitted
with a great deal of power. Also note the large KEEP OUT signs near
radio transmission towers for local radio stations. Standing within 10 ft
of an AM transmitter working at 540 kHz would bring on disaster. Simply holding (not to be tried!) a fluorescent bulb near the tower could
make it light up due to the excitation of the molecules inside the bulb.
In total, therefore, treat any situation with high ac voltages or currents, high-energy dc levels, and high frequencies with added care.
Bulb Savers
Ever since the invention of the light bulb, consumers have clamored for
ways to extend the life of a bulb. I can remember the days when I was
taught to always turn a light off when leaving a room and not to play
with a light switch because it cost us a penny (at a time when a penny
had some real value) every time I turned the switch on and off. Through
advanced design we now have bulbs that are guaranteed to last a number of years. They cost more, but there is no need to replace the bulb as
often, and over time there is a financial savings. For some of us it is
simply a matter of having to pay so much for a single bulb.
For interest sake, I measured the cold dc resistance of a standard
60-W bulb and found it to be about 14 . Forgetting any inductive
effects due to the filament and wire, this would mean a current of
120 V/14 8.6 A when the light is first turned on. This is a fairly
heavy current for the filament to absorb when you consider that the normal operating current is 60 W/120 V 0.5 A. Fortunately, it lasts for
only a few milliseconds, as shown in Fig. 13.73(a), before the bulb
heats up, causing the filament resistance to quickly increase and cut the
current down to reasonable levels. However, over time, hitting the bulb
with 8.6 A every time you turn the switch on will take its toll on the filament, and eventually the filament will simply surrender its natural
characteristics and open up. You can easily tell if a bulb is bad by simply shaking it and listening for the clinking sound of the broken filament hitting the bulb. Assuming an initial current of 8.6 A for a single
bulb, if the light switch controlled four 60-W bulbs in the same room,
the surge current through the switch could be as high as 4(8.6 A) 34.4 A as shown in Fig. 13.73(b), which probably exceeds the rating of
the breaker (typically 20 A) for the circuit. However, the saving grace is
that it lasts for only a few milliseconds, and circuit breakers are not
designed to react that quickly. Even the GFI safety breakers in the bathroom are typically rated at a 5-ms response time. However, when you
look at the big picture and imagine all these spikes on the line generated throughout a residential community, it is certainly a problem that
the power company has to deal with on a continuing basis.
One way to suppress this surge current is to place an inductor in
series with the bulb to choke out the spikes down the line. This method,
in fact, leads to one way of extending the life of a light bulb through the
use of dimmers. Any well-designed dimmer (such as the one described
in Chapter 12) has an inductor in the line to suppress current surges.
The results are both an extended life for the bulb and the ability to control the power level. Left on in the full voltage position, the switch
could be used as a regular switch and the life of the bulb could be
Ibulb (A)
8.6
0.5
0
1
2
3
4
5
6
7 t (ms)
(a)
IT (A), four parallel 60-W bulbs
34.4
2
0
0.5
1
1.5
2
t (s)
250 ms
(b)
FIG 13.73
Surge currents: (a) single 60-W bulb; (b) four
parallel 60-W bulbs.
546

SINUSOIDAL ALTERNATING WAVEFORMS
V (volts)
V (volts)
Vapplied
170
170
120 V rms
ac voltage
85
85
120°
180°
0
30°
150°
Vbulb
360°
θ
0
–85
–85
–170
–170
(a)
180°
360°
t
90°
Turn-on
Dimmer ensures
bulb will not
turn on when
applied voltage is
near its maximum.
(b)
FIG. 13.74
Turn-on voltage: (a) equal to or greater than one-half the peak value; (b) when
a dimmer is used.
extended. In fact, many dimmers now use triacs designed to turn on
only when the applied voltage passes through zero. If we look at the full
sine wave of Fig. 13.74(a), we find that the voltage is at least half of its
maximum value of 85 V for a full two-thirds of each cycle, or about
67% of the time. The chances, therefore, of your turning on a light bulb
with at least 85 V on the line is far better than 2 to 1, so you can expect
the current for a 60-W light bulb to be at least 85 V/14 6 A 67%
of the time, which exceeds the rated 0.5-A rated value by 1100%. If we
use a dimmer with a triac designed to turn on only when the applied
voltage passes through zero or shortly thereafter, as shown in Fig.
(a)
120-V rms line
+
Id
+0.7 V – +0.7 V
Ilamp
(for positive region
of applied voltage)
High resistance
Id
170 V
peak
+
168.6 V
Lamp
–
–
Low
resistance
Open circuits for
positive region of
applied voltage
0
High resistance
(diode open circuit)
Return
(b)
+ Vd –
Vd
(c)
FIG. 13.75
Bulb saver: (a) external appearance; (b) basic operation; (c) diode
characteristics at high current levels.
COMPUTER ANALYSIS
13.74(b), the applied voltage will increase from about zero volts, giving
the bulb time to warm up before the full voltage is applied.
Another commercial offering to extend the life of light bulbs is the
smaller circular disc shown in Fig. 13.75(a) which is inserted between
the bulb and the holder. Contacts are provided on both sides to permit
conduction through the simple diode network shown in Fig. 13.75(b).
You may recall from an earlier chapter that the voltage across diodes in
the on state is 0.7 V as shown for each diode in Fig. 13.75(b) for the
positive portion of the input voltage. The result is that the voltage to the
bulb is reduced by about 1.4 V throughout the cycle, reducing the
power delivered to the bulb. For most situations the reduced lighting is
not a problem, and the bulb will last longer simply because it is not
pressed to work at full output. However, the real saving in the device is
the manner in which it could help suppress the surge currents through
the light bulb. The true characteristics of a diode are shown in Fig.
13.75(c) for the full range of currents through the diode. For most
applications in electronic circuits, the vertical region is employed. For
excessive currents the diode characteristics flatten out as shown. This
region is characterized as having a large resistance (compared to very
small resistance of the vertical region) which will come into play when
the bulb is first turned on. In other words, when the bulb is first turned
on, the current will be so high that the diode will enter its high resistance region and by Ohm’s law will limit the surge current—thereby
extending the life of the bulb. The two diodes facing the other way are
for the negative portion of the supply voltage.
New methods of extending the life of bulbs hit the marketplace every
day. All in all, however, one guaranteed way to extend the life of your
bulbs is to return to the old philosophy of turning lights off when you
leave a room, and “Don’t play with the light switch!”
13.10 COMPUTER ANALYSIS
PSpice (Windows)
Schematics offer a variety of ac voltage and current sources. However,
for the purposes of this text, the voltage source VSIN and the current
source ISIN are the most appropriate because they have a list of attributes that will cover most areas of normal interest for sinusoidal networks. Under the library, SOURCE.slb, a number of others are listed,
but they don’t have the full range of the above or they are dedicated to
only one type of analysis. On occasion, ISRC will be used because it
has an arrow symbol like that appearing in the text, and it can be used
for dc, ac, and some transient analyses. The symbol for ISIN is simply
a sine wave which utilizes the plus-and-minus sign to indicate direction.
The sources VAC, IAC, VSRC, and ISRC are fine if the magnitude and
phase of a specific quantity are desired or if a transient plot is against
frequency. However, they will not provide a transient response against
time even if the frequency and transient information are provided under
Analysis.
For all of the sinusoidal sources, the magnitude entered and read is
the peak value of the waveform and not the rms value. This will become
clear when a plot of a quantity is desired and the magnitude calculated
by PSpice (Windows) is the peak value of the transient response. However, for a purely steady-state ac response, the magnitude provided can

547
548

SINUSOIDAL ALTERNATING WAVEFORMS
be the rms value, and the output read as the rms value. Only when a plot
is desired will it be clear that PSpice is accepting every ac magnitude as
the peak value of the waveform. Of course, the phase angle is the same
whether the magnitude is the peak or the rms value.
A number of default values are set by PSpice if values are entered
for specific attributes of the source. If not specified, DC and AC values
are defaulted to 0, and Transient values default to the DC value. When
using VSIN, always specify VOFF as 0 V (unless a specific value is
part of the analysis), provide both the AC and the VAMPL values at the
same level, and provide the PHASE angle associated with the source.
The TD (time delay), DF (damping factor)